`int x^2/(x^2+1)^2 dx`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the given indefinite integral, hence, you may start by addig and subtracting 1 to numerator, such that:

`int x^2/((x^2 + 1)^2) dx = int (x^2 + 1 - 1)/((x^2 + 1)^2) dx`

You need to split the integral into two simpler integrals, using the property of linearity of integrals, such that:

`int x^2/((x^2 + 1)^2) dx = int (x^2 + 1)/((x^2 + 1)^2) dx - int 1/((x^2 + 1)^2) dx`

Reducing duplicate factors yields:

`int x^2/((x^2 + 1)^2) dx = int 1/(x^2 + 1)dx - int 1/((x^2 + 1)^2) dx`

`int x^2/((x^2 + 1)^2) dx = tan^(-1) x + int 1/((x^2 + 1)^2) dx`

You need to evaluate the indefinite integral `int 1/((x^2 + 1)^2) dx` , hence, you need to come up with the substitution, such that:

`x = tan y => dx = (tan^2 y + 1)dy`

Changing the variable, yields:

`int 1/((x^2 + 1)^2) dx = int ((tan^2 y + 1)dy)/((tan^2 y + 1)^2)`

You need to reduce duplicate factors, such that:

`int ((tan^2 y + 1)dy)/((tan^2 y + 1)^2) = int (dy)/(tan^2 y + 1)`

Using the trigonometric identity `tan^2 y + 1 = 1/(cos^2 y)` yields:

`int (dy)/(tan^2 y + 1) = int (dy)/(1/(cos^2 y)) = int cos^2 y dy`

Using the half angle trigonometric identity `cos^2 y = (1 + cos 2y)/2` , yields:

`int cos^2 y dy = int (1 + cos 2y)/2 dy`

`int cos^2 y dy = (1/2) int (1 + cos 2y)dy`

`int cos^2 y dy = (1/2) int dy + (1/2) cos 2y dy`

`int cos^2 y dy = (1/2)(y + (sin2y)/2)`

You need to replace `tan^(-1) x` for `y` , such that:

`int 1/((x^2 + 1)^2) dx = (1/2)(tan^(-1) x + (sin2tan^(-1) x)/2) + c`

`int x^2/((x^2 + 1)^2) dx = (3/2)tan^(-1) x + (1/2)((sin2tan^(-1) x)/2) + c`

Hence, evaluating the given indefinite integral, using the integration by substitution, yields `int x^2/((x^2 + 1)^2) dx = (3/2)tan^(-1) x + (1/2)((sin(2tan^(-1) x))/2) + c.`

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nikolao | High School Teacher | (Level 1) Adjunct Educator

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int  x^2/(x^2+1)^2  dx

To solve this  let’s use    x=tan(theta)

x=tan(theta)    so           dx=sec^2 theta d(theta)

int  x^2/(x^2+1)^2  dx

= int (tan^2 (theta) . sec^2 (theta))/(1+tan^2 theta)^2   d(theta)

= int tan^2 theta . cos^2 theta d(theta)

=int sin^2 theta d(theta)

=(1/2)int (1-cos2(theta)) d(theta)

=(1/2)(theta)-(1/4) sin(2theta)

tan (theta) = x   ,  sin(2theta) = 2x/(1+x^2)

int x^2/(x^2+1)^2  dx = (1/2) tan^-1 x – x/2(1+x^2) +c

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