What is the integral of (1+x^2)^{-2} ?

The integral of (1+x^2)^{-2} is 1/2 (arctan(x)+x/(1+x^2))+C

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integral of (1+x^2)^{-2}

\int (1 + x^{2})^{- 2} d x

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

Solution steps

\int (1 + x^{2})^{- 2} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

u = arctan (x)

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

Simplify

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

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

Add a constant to the solution

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

(x^{6})^{^{'}}

\frac{d}{d x} (9 + x^{2})

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\int x^{3} (5 + x^{4})^{11} d x

\int_{x^{3}}^{x^{4}} (2 t - 1)^{3} d t

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