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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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\int sec^{2} (5 x) 2^{t a n (5 x)} d x

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