Mixing
Finding the definite integral $int_1^e frac{dx}{xsqrt{1+ln^2x}}$
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So I have the following problem:
$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$
Can somebody comfirm that the integral of this is
$$ln|sqrt{1+ln^2x}+ ln x|+C$$
and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814
Does anyone else got the same anwser?
calculus integration definite-integrals closed-form
asked 2 days ago
Student123
313
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Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
5
down vote
favorite
So I have the following problem:
$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$
Can somebody comfirm that the integral of this is
$$ln|sqrt{1+ln^2x}+ ln x|+C$$
and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814
Does anyone else got the same anwser?
calculus integration definite-integrals closed-form
asked 2 days ago
Student123
313
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
1
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
So I have the following problem:
$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$
Can somebody comfirm that the integral of this is
$$ln|sqrt{1+ln^2x}+ ln x|+C$$
and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814
Does anyone else got the same anwser?
calculus integration definite-integrals closed-form
asked 2 days ago
Student123
313
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So I have the following problem:
$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$
Can somebody comfirm that the integral of this is
$$ln|sqrt{1+ln^2x}+ ln x|+C$$
and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814
Does anyone else got the same anwser?
calculus integration definite-integrals closed-form
calculus integration definite-integrals closed-form
asked 2 days ago
Student123
313
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Student123
313
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
asked 2 days ago
Student123
313
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Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
Student123
313
asked 2 days ago
Student123
313
313
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Student123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
1
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago
add a comment |
3
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
1
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago
3
3
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
1
1
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$
answered 2 days ago
Olivier Oloa
107k17175293
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
add a comment |
up vote
2
down vote
once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED
Remember that $log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.
answered yesterday
clathratus
1,729219
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$
answered 2 days ago
Olivier Oloa
107k17175293
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
add a comment |
up vote
5
down vote
Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$
answered 2 days ago
Olivier Oloa
107k17175293
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$
answered 2 days ago
Olivier Oloa
107k17175293
Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$
answered 2 days ago
Olivier Oloa
107k17175293
answered 2 days ago
Olivier Oloa
107k17175293
answered 2 days ago
Olivier Oloa
107k17175293
answered 2 days ago
Olivier Oloa
107k17175293
107k17175293
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
add a comment |
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
– Student123
19 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
@Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
– Olivier Oloa
18 hours ago
add a comment |
up vote
2
down vote
once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED
Remember that $log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.
answered yesterday
clathratus
1,729219
add a comment |
up vote
2
down vote
once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED
Remember that $log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.
answered yesterday
clathratus
1,729219
add a comment |
up vote
2
down vote
up vote
2
down vote
once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED
Remember that $log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.
answered yesterday
clathratus
1,729219
once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED
Remember that $log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.
answered yesterday
clathratus
1,729219
edited yesterday
answered yesterday
clathratus
1,729219
answered yesterday
clathratus
1,729219
answered yesterday
clathratus
1,729219
1,729219
add a comment |
add a comment |
Student123 is a new contributor. Be nice, and check out our Code of Conduct.
Student123 is a new contributor. Be nice, and check out our Code of Conduct.
Student123 is a new contributor. Be nice, and check out our Code of Conduct.
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3
Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
2 days ago
1
You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
2 days ago
Differentiate your ans, with respect to x.
– John Nash
2 days ago