What is the integral of (8x^7)/((x^8+1)^2) ?

The integral of (8x^7)/((x^8+1)^2) is -1/(x^8+1)+C

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integral of (8x^7)/((x^8+1)^2)

\int \frac{8 x ^{7}}{( x ^{8} + 1 ) ^{2}} d x

- \frac{1}{x ^{8} + 1} + C

Solution steps

\int \frac{8 x ^{7}}{( x ^{8} + 1 ) ^{2}} 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 ^{7}}{( x ^{8} + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{8 u ^{2}} d u

= 8 \cdot \int \frac{1}{8 u ^{2}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x^{8} + 1

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

Simplify

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

= - \frac{1}{x ^{8} + 1}

Add a constant to the solution

= - \frac{1}{x ^{8} + 1} + C

\frac{d}{d x} (\frac{d x}{x})

(1 + 3 t)^{2}

\frac{d}{d t} (t sin (t) + cos (t))

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

(1 + x)^{4}

What is the integral of (8x^7)/((x^8+1)^2) ?
The integral of (8x^7)/((x^8+1)^2) is -1/(x^8+1)+C