Mixing
integral with parameter and 2 denominators
$begingroup$
Hello I want to solve following integral
$displaystyleint frac{x^22t}{(1+x^2)(1+t^2x^2)},dx=-dfrac{2arctan(tx)-t arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
integration
edited Jan 11 at 20:46
Bernard
120k740116
asked Jan 11 at 20:32
tim123tim123
173
$endgroup$
add a comment |
$begingroup$
Hello I want to solve following integral
$displaystyleint frac{x^22t}{(1+x^2)(1+t^2x^2)},dx=-dfrac{2arctan(tx)-t arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
integration
edited Jan 11 at 20:46
Bernard
120k740116
asked Jan 11 at 20:32
tim123tim123
173
$endgroup$
2
$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33
add a comment |
$begingroup$
Hello I want to solve following integral
$displaystyleint frac{x^22t}{(1+x^2)(1+t^2x^2)},dx=-dfrac{2arctan(tx)-t arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
integration
edited Jan 11 at 20:46
Bernard
120k740116
asked Jan 11 at 20:32
tim123tim123
173
$endgroup$
Hello I want to solve following integral
$displaystyleint frac{x^22t}{(1+x^2)(1+t^2x^2)},dx=-dfrac{2arctan(tx)-t arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
integration
integration
edited Jan 11 at 20:46
Bernard
120k740116
asked Jan 11 at 20:32
tim123tim123
173
edited Jan 11 at 20:46
Bernard
120k740116
asked Jan 11 at 20:32
tim123tim123
173
edited Jan 11 at 20:46
Bernard
120k740116
edited Jan 11 at 20:46
Bernard
120k740116
edited Jan 11 at 20:46
Bernard
120k740116
120k740116
asked Jan 11 at 20:32
tim123tim123
173
asked Jan 11 at 20:32
tim123tim123
173
asked Jan 11 at 20:32
tim123tim123
173
173
2
$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33
add a comment |
2
$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33
2
2
$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33
$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Partial fraction decomposition gives$$frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}=frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$begin{align*}mathfrak{I} & =2tintmathrm dx,left[frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}right]\ & =frac {2t}{1-t^2}intfrac {mathrm dx}{1+t^2x^2}-frac {2t}{1-t^2}intfrac {mathrm dx}{1+x^2}end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$frac {x^2}{(1+x^2)(1+t^2x^2)}=frac {Ax+B}{1+x^2}+frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$begin{align*}-1 & =(Ax+B)(1-t^2)\ & =Ax(1-t^2)+B(1-t^2)end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-tfrac 1{t^2}$ to get that $C=0$ and $D=tfrac 1{1-t^2}$.$$frac {x^2}{(1+x^2)(1+t^2x^2)}color{red}{=frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
add a comment |
$begingroup$
Hint: Write your integrand in the form
$$frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
Your Answer
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2 Answers
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$begingroup$
Partial fraction decomposition gives$$frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}=frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$begin{align*}mathfrak{I} & =2tintmathrm dx,left[frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}right]\ & =frac {2t}{1-t^2}intfrac {mathrm dx}{1+t^2x^2}-frac {2t}{1-t^2}intfrac {mathrm dx}{1+x^2}end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$frac {x^2}{(1+x^2)(1+t^2x^2)}=frac {Ax+B}{1+x^2}+frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$begin{align*}-1 & =(Ax+B)(1-t^2)\ & =Ax(1-t^2)+B(1-t^2)end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-tfrac 1{t^2}$ to get that $C=0$ and $D=tfrac 1{1-t^2}$.$$frac {x^2}{(1+x^2)(1+t^2x^2)}color{red}{=frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
add a comment |
$begingroup$
Partial fraction decomposition gives$$frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}=frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$begin{align*}mathfrak{I} & =2tintmathrm dx,left[frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}right]\ & =frac {2t}{1-t^2}intfrac {mathrm dx}{1+t^2x^2}-frac {2t}{1-t^2}intfrac {mathrm dx}{1+x^2}end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$frac {x^2}{(1+x^2)(1+t^2x^2)}=frac {Ax+B}{1+x^2}+frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$begin{align*}-1 & =(Ax+B)(1-t^2)\ & =Ax(1-t^2)+B(1-t^2)end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-tfrac 1{t^2}$ to get that $C=0$ and $D=tfrac 1{1-t^2}$.$$frac {x^2}{(1+x^2)(1+t^2x^2)}color{red}{=frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
add a comment |
$begingroup$
Partial fraction decomposition gives$$frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}=frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$begin{align*}mathfrak{I} & =2tintmathrm dx,left[frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}right]\ & =frac {2t}{1-t^2}intfrac {mathrm dx}{1+t^2x^2}-frac {2t}{1-t^2}intfrac {mathrm dx}{1+x^2}end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$frac {x^2}{(1+x^2)(1+t^2x^2)}=frac {Ax+B}{1+x^2}+frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$begin{align*}-1 & =(Ax+B)(1-t^2)\ & =Ax(1-t^2)+B(1-t^2)end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-tfrac 1{t^2}$ to get that $C=0$ and $D=tfrac 1{1-t^2}$.$$frac {x^2}{(1+x^2)(1+t^2x^2)}color{red}{=frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
$endgroup$
Partial fraction decomposition gives$$frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}=frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$begin{align*}mathfrak{I} & =2tintmathrm dx,left[frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}right]\ & =frac {2t}{1-t^2}intfrac {mathrm dx}{1+t^2x^2}-frac {2t}{1-t^2}intfrac {mathrm dx}{1+x^2}end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$frac {x^2}{(1+x^2)(1+t^2x^2)}=frac {Ax+B}{1+x^2}+frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$begin{align*}-1 & =(Ax+B)(1-t^2)\ & =Ax(1-t^2)+B(1-t^2)end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-tfrac 1{t^2}$ to get that $C=0$ and $D=tfrac 1{1-t^2}$.$$frac {x^2}{(1+x^2)(1+t^2x^2)}color{red}{=frac 1{(1-t^2)(1+t^2x^2)}-frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
edited Jan 12 at 4:09
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
answered Jan 11 at 20:45
Frank W.Frank W.
3,5731321
3,5731321
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
add a comment |
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
I do not really understand how I do come to this partial fraction decomposition, since i do not have a pole.
$endgroup$
– tim123
Jan 11 at 21:17
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
$begingroup$
@tim123 I've added an explanation. If you have any more questions, feel free to ask!
$endgroup$
– Frank W.
Jan 12 at 4:09
add a comment |
$begingroup$
Hint: Write your integrand in the form
$$frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
Hint: Write your integrand in the form
$$frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
Hint: Write your integrand in the form
$$frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
Hint: Write your integrand in the form
$$frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 11 at 20:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
75.1k42865
add a comment |
add a comment |
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$begingroup$
Partial fraction decomposition is your friend.
$endgroup$
– Don Thousand
Jan 11 at 20:33