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

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

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

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

- \frac{1}{x ^{4} + 1} + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = x^{4} + 1

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

Simplify

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

= - \frac{1}{x ^{4} + 1}

Add a constant to the solution

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

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