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

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

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integral of (4x^3)/((x^2+1)^3)

\int \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

Apply u-substitution

= 4 \cdot \int \frac{u - 1}{2 u ^{3}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{u - 1}{u ^{3}} d u

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{3}} 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

= 4 \cdot \frac{1}{2} (\int \frac{1}{u ^{2}} d u - \int \frac{1}{u ^{3}} d u)

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

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

= 4 \cdot \frac{1}{2} (- \frac{1}{u} - (- \frac{1}{2 u ^{2}}))

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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