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

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

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

integral

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

- \frac{1}{4 ( x ^{2} + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{2 u ^{3}} 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 \frac{1}{u ^{3}} d u

Apply the Power Rule

= \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{2} + 1

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

Simplify

\frac{1}{2} (- \frac{1}{2 ( x ^{2} + 1 ) ^{2}}) : - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

= - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( x ^{2} + 1 ) ^{2}} + C

in t e g r a l \frac{( x + 8 )}{( x + 7 )}

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