What is the integral of (8x^3)/((x^4+1)^3) ?

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

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integral of (8x^3)/((x^4+1)^3)

\int \frac{8 x ^{3}}{( x ^{4} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{8 x ^{3}}{( x ^{4} + 1 ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{3}}{( x ^{4} + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{3}} d u

= 8 \cdot \int \frac{1}{4 u ^{3}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{4} + 1

= 8 \cdot \frac{1}{4} (- \frac{1}{2 ( x ^{4} + 1 ) ^{2}})

Simplify

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

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

Add a constant to the solution

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

\int \frac{ln ( 10 x )}{x} d x

y^{^{''}} + 12 y^{^{'}} + 24 y = 4381 e^{2 x} cos (13 x)

\frac{2 s}{( s + 3 ) ^{2}}

\int e^{- s t} sin (b t) d t

y = 1 - x^{2}, y = x^{2} - 1

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