What is the integral of pi(1/(1+x^2))^2 ?

The integral of pi(1/(1+x^2))^2 is pi/2 (arctan(x)+x/(1+x^2))+C

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integral of pi(1/(1+x^2))^2

\int π (\frac{1}{1 + x ^{2}})^{2} d x

\frac{π}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + C

Solution steps

\int π (\frac{1}{1 + x ^{2}})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int (\frac{1}{1 + x ^{2}})^{2} d x

Simplify

(\frac{1}{1 + x ^{2}})^{2} : \frac{1}{( 1 + x ^{2} ) ^{2}}

= π \cdot \int \frac{1}{( 1 + x ^{2} ) ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

= π \cdot \int \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= π \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= π \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arctan (x)

= π \frac{1}{2} (arctan (x) + \frac{1}{2} sin (2 arctan (x)))

Simplify

π \frac{1}{2} (arctan (x) + \frac{1}{2} sin (2 arctan (x))) : \frac{π}{2} (arctan (x) + \frac{x}{1 + x ^{2}})

= \frac{π}{2} (arctan (x) + \frac{x}{1 + x ^{2}})

Add a constant to the solution

= \frac{π}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + C

\int_{0}^{1} \frac{2 t}{1 + t ^{2}} d t

y = sin (4 x)

\frac{d}{d x} (\frac{10 x + x ^{2}}{( 5 + x ) ^{2}})

\int r^{3} e^{- 2 r} d r

\frac{\partial}{\partial x} ((\frac{x}{y})^{z})

What is the integral of pi(1/(1+x^2))^2 ?
The integral of pi(1/(1+x^2))^2 is pi/2 (arctan(x)+x/(1+x^2))+C