Mixing
Evaluate $int frac{x^2}{x-1} ,dx$
$begingroup$
Evaluate $int frac{x^2}{x-1} ,dx$
(A) $2x^2+x+ln|x-1|+C$
(B) $frac{x^2}{2}+x+ln|x+1|+C$
(C) $frac{x^2}{2}+x+ln|x-1|+C$
(D) $x^2+x+ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$int frac{(u+1)^2}{u},du\
=int u +2+frac{1}{u},du\
=frac{u^2}{2}+2u+ln|u|+C\
=frac{(x-1)^2}{2}+2(x-1)+ln|x-1|+C$
Simplify :
$frac{x^2-3}{2}+x+ln|x-1|+C$
It's not on the option.
calculus integration
asked Feb 1 at 3:09
airlanggaairlangga
775
$endgroup$
add a comment |
$begingroup$
Evaluate $int frac{x^2}{x-1} ,dx$
(A) $2x^2+x+ln|x-1|+C$
(B) $frac{x^2}{2}+x+ln|x+1|+C$
(C) $frac{x^2}{2}+x+ln|x-1|+C$
(D) $x^2+x+ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$int frac{(u+1)^2}{u},du\
=int u +2+frac{1}{u},du\
=frac{u^2}{2}+2u+ln|u|+C\
=frac{(x-1)^2}{2}+2(x-1)+ln|x-1|+C$
Simplify :
$frac{x^2-3}{2}+x+ln|x-1|+C$
It's not on the option.
calculus integration
asked Feb 1 at 3:09
airlanggaairlangga
775
$endgroup$
3
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16
add a comment |
$begingroup$
Evaluate $int frac{x^2}{x-1} ,dx$
(A) $2x^2+x+ln|x-1|+C$
(B) $frac{x^2}{2}+x+ln|x+1|+C$
(C) $frac{x^2}{2}+x+ln|x-1|+C$
(D) $x^2+x+ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$int frac{(u+1)^2}{u},du\
=int u +2+frac{1}{u},du\
=frac{u^2}{2}+2u+ln|u|+C\
=frac{(x-1)^2}{2}+2(x-1)+ln|x-1|+C$
Simplify :
$frac{x^2-3}{2}+x+ln|x-1|+C$
It's not on the option.
calculus integration
asked Feb 1 at 3:09
airlanggaairlangga
775
$endgroup$
Evaluate $int frac{x^2}{x-1} ,dx$
(A) $2x^2+x+ln|x-1|+C$
(B) $frac{x^2}{2}+x+ln|x+1|+C$
(C) $frac{x^2}{2}+x+ln|x-1|+C$
(D) $x^2+x+ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$int frac{(u+1)^2}{u},du\
=int u +2+frac{1}{u},du\
=frac{u^2}{2}+2u+ln|u|+C\
=frac{(x-1)^2}{2}+2(x-1)+ln|x-1|+C$
Simplify :
$frac{x^2-3}{2}+x+ln|x-1|+C$
It's not on the option.
calculus integration
calculus integration
asked Feb 1 at 3:09
airlanggaairlangga
775
asked Feb 1 at 3:09
airlanggaairlangga
775
asked Feb 1 at 3:09
airlanggaairlangga
775
asked Feb 1 at 3:09
airlanggaairlangga
775
asked Feb 1 at 3:09
airlanggaairlangga
775
775
3
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16
add a comment |
3
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16
3
3
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$frac{x^2-3}{2}+x+ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-frac32$, which is fine since the difference is constant.
Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.
As a side note, it isn't really correct to say that $intfrac1x dx=ln|x|+C$, because of the discontinuity of $frac1x$ at $0$. It's a bit more nuanced than that:
$$
intfrac1x dx =
begin{cases}
ln|x| + C_1, & text{if $x > 0$} \
ln|x| + C_2, & text{if $x < 0$}
end{cases}$$
So in that sense, even the given answers are not entirely right.
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
Your answer is right except for the constant I changed.
$$frac{x^2-3}{2}+x+ln|x-1|+C_1 = frac{x^2}{2}+x+ln|x-1|+C_1-frac{3}{2} = frac{x^2}{2}+x+ln|x-1|+C$$
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
$endgroup$
add a comment |
$begingroup$
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
begin{align}
int frac{x^2}{x-1} ,dx
&=int frac{x^2-1+1}{x-1} ,dx\
&=int left(frac{x^2-1}{x-1}+frac{1}{x-1}right) ,dx\
&=int frac{(x-1)(x+1)}{x-1},dx+intfrac{1}{x-1} ,dx\
&=int left(x+1right),dx+intfrac{1}{x-1}frac{d}{dx}left(x-1right) ,dx\
&=frac{x^2}{2}+x+intfrac{1}{x-1},dleft(x-1right)\
&=frac{x^2}{2}+x+ln{left|x-1right|}+C.\
end{align}
$$
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
$endgroup$
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$frac{x^2-3}{2}+x+ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-frac32$, which is fine since the difference is constant.
Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.
As a side note, it isn't really correct to say that $intfrac1x dx=ln|x|+C$, because of the discontinuity of $frac1x$ at $0$. It's a bit more nuanced than that:
$$
intfrac1x dx =
begin{cases}
ln|x| + C_1, & text{if $x > 0$} \
ln|x| + C_2, & text{if $x < 0$}
end{cases}$$
So in that sense, even the given answers are not entirely right.
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$frac{x^2-3}{2}+x+ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-frac32$, which is fine since the difference is constant.
Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.
As a side note, it isn't really correct to say that $intfrac1x dx=ln|x|+C$, because of the discontinuity of $frac1x$ at $0$. It's a bit more nuanced than that:
$$
intfrac1x dx =
begin{cases}
ln|x| + C_1, & text{if $x > 0$} \
ln|x| + C_2, & text{if $x < 0$}
end{cases}$$
So in that sense, even the given answers are not entirely right.
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$frac{x^2-3}{2}+x+ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-frac32$, which is fine since the difference is constant.
Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.
As a side note, it isn't really correct to say that $intfrac1x dx=ln|x|+C$, because of the discontinuity of $frac1x$ at $0$. It's a bit more nuanced than that:
$$
intfrac1x dx =
begin{cases}
ln|x| + C_1, & text{if $x > 0$} \
ln|x| + C_2, & text{if $x < 0$}
end{cases}$$
So in that sense, even the given answers are not entirely right.
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
$endgroup$
You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$frac{x^2-3}{2}+x+ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-frac32$, which is fine since the difference is constant.
Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.
As a side note, it isn't really correct to say that $intfrac1x dx=ln|x|+C$, because of the discontinuity of $frac1x$ at $0$. It's a bit more nuanced than that:
$$
intfrac1x dx =
begin{cases}
ln|x| + C_1, & text{if $x > 0$} \
ln|x| + C_2, & text{if $x < 0$}
end{cases}$$
So in that sense, even the given answers are not entirely right.
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
answered Feb 1 at 4:02
ThéophileThéophile
20.4k13047
20.4k13047
add a comment |
add a comment |
$begingroup$
Your answer is right except for the constant I changed.
$$frac{x^2-3}{2}+x+ln|x-1|+C_1 = frac{x^2}{2}+x+ln|x-1|+C_1-frac{3}{2} = frac{x^2}{2}+x+ln|x-1|+C$$
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
$endgroup$
add a comment |
$begingroup$
Your answer is right except for the constant I changed.
$$frac{x^2-3}{2}+x+ln|x-1|+C_1 = frac{x^2}{2}+x+ln|x-1|+C_1-frac{3}{2} = frac{x^2}{2}+x+ln|x-1|+C$$
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
$endgroup$
add a comment |
$begingroup$
Your answer is right except for the constant I changed.
$$frac{x^2-3}{2}+x+ln|x-1|+C_1 = frac{x^2}{2}+x+ln|x-1|+C_1-frac{3}{2} = frac{x^2}{2}+x+ln|x-1|+C$$
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
$endgroup$
Your answer is right except for the constant I changed.
$$frac{x^2-3}{2}+x+ln|x-1|+C_1 = frac{x^2}{2}+x+ln|x-1|+C_1-frac{3}{2} = frac{x^2}{2}+x+ln|x-1|+C$$
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
answered Feb 1 at 4:29
Pranita GuptaPranita Gupta
123112
123112
add a comment |
add a comment |
$begingroup$
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
begin{align}
int frac{x^2}{x-1} ,dx
&=int frac{x^2-1+1}{x-1} ,dx\
&=int left(frac{x^2-1}{x-1}+frac{1}{x-1}right) ,dx\
&=int frac{(x-1)(x+1)}{x-1},dx+intfrac{1}{x-1} ,dx\
&=int left(x+1right),dx+intfrac{1}{x-1}frac{d}{dx}left(x-1right) ,dx\
&=frac{x^2}{2}+x+intfrac{1}{x-1},dleft(x-1right)\
&=frac{x^2}{2}+x+ln{left|x-1right|}+C.\
end{align}
$$
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
$endgroup$
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
add a comment |
$begingroup$
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
begin{align}
int frac{x^2}{x-1} ,dx
&=int frac{x^2-1+1}{x-1} ,dx\
&=int left(frac{x^2-1}{x-1}+frac{1}{x-1}right) ,dx\
&=int frac{(x-1)(x+1)}{x-1},dx+intfrac{1}{x-1} ,dx\
&=int left(x+1right),dx+intfrac{1}{x-1}frac{d}{dx}left(x-1right) ,dx\
&=frac{x^2}{2}+x+intfrac{1}{x-1},dleft(x-1right)\
&=frac{x^2}{2}+x+ln{left|x-1right|}+C.\
end{align}
$$
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
$endgroup$
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
add a comment |
$begingroup$
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
begin{align}
int frac{x^2}{x-1} ,dx
&=int frac{x^2-1+1}{x-1} ,dx\
&=int left(frac{x^2-1}{x-1}+frac{1}{x-1}right) ,dx\
&=int frac{(x-1)(x+1)}{x-1},dx+intfrac{1}{x-1} ,dx\
&=int left(x+1right),dx+intfrac{1}{x-1}frac{d}{dx}left(x-1right) ,dx\
&=frac{x^2}{2}+x+intfrac{1}{x-1},dleft(x-1right)\
&=frac{x^2}{2}+x+ln{left|x-1right|}+C.\
end{align}
$$
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
$endgroup$
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
begin{align}
int frac{x^2}{x-1} ,dx
&=int frac{x^2-1+1}{x-1} ,dx\
&=int left(frac{x^2-1}{x-1}+frac{1}{x-1}right) ,dx\
&=int frac{(x-1)(x+1)}{x-1},dx+intfrac{1}{x-1} ,dx\
&=int left(x+1right),dx+intfrac{1}{x-1}frac{d}{dx}left(x-1right) ,dx\
&=frac{x^2}{2}+x+intfrac{1}{x-1},dleft(x-1right)\
&=frac{x^2}{2}+x+ln{left|x-1right|}+C.\
end{align}
$$
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
edited Feb 1 at 4:04
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
answered Feb 1 at 3:36
Michael RybkinMichael Rybkin
4,239422
4,239422
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
add a comment |
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
1
1
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
$begingroup$
The OP has already done all of this; the question concerns the constant at the end.
$endgroup$
– Théophile
Feb 1 at 20:22
add a comment |
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3
$begingroup$
Your answer is $frac{x^2}2+x+ln|x-1|+(C-frac32)$. Can you resolve the matter from here?
$endgroup$
– David
Feb 1 at 3:12
$begingroup$
@David so, is it C?
$endgroup$
– airlangga
Feb 1 at 3:15
$begingroup$
yea you can break up the $frac{x^2-3}{2}$
$endgroup$
– user29418
Feb 1 at 3:16