What is the integral of (x^2)/((1+x^2)^3) ?

The integral of (x^2)/((1+x^2)^3) is 1/8 (arctan(x)-1/4 sin(4arctan(x)))+C

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integral of (x^2)/((1+x^2)^3)

\int \frac{x ^{2}}{( 1 + x ^{2} ) ^{3}} d x

\frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x))) + C

Solution steps

\int \frac{x ^{2}}{( 1 + x ^{2} ) ^{3}} d x

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{sec ^{6} ( u ) cos ^{2} ( u )} d u

= \int \frac{tan ^{2} ( u )}{sec ^{6} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 1 - cos (4 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}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arctan (x)

= \frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x)))

Add a constant to the solution

= \frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x))) + C

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

\int \frac{5 x}{x ^{2} + 3} d x

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

\int \frac{x ^{2} - 1}{x ^{3} + 1} d x

\int \frac{x}{16 - 4 x ^{2}} d x

What is the integral of (x^2)/((1+x^2)^3) ?
The integral of (x^2)/((1+x^2)^3) is 1/8 (arctan(x)-1/4 sin(4arctan(x)))+C