What is the integral of (10x)/((x^2+1)^6) ?

The integral of (10x)/((x^2+1)^6) is -1/((x^2+1)^5)+C

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integral of (10x)/((x^2+1)^6)

\int \frac{10 x}{( x ^{2} + 1 ) ^{6}} d x

- \frac{1}{( x ^{2} + 1 ) ^{5}} + C

Solution steps

\int \frac{10 x}{( x ^{2} + 1 ) ^{6}} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{6}} d x

Apply u-substitution

= 10 \cdot \int \frac{1}{2 u ^{6}} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

= 10 \cdot \frac{1}{2} (- \frac{1}{5 u ^{5}})

Substitute back

u = x^{2} + 1

= 10 \cdot \frac{1}{2} (- \frac{1}{5 ( x ^{2} + 1 ) ^{5}})

Simplify

10 \cdot \frac{1}{2} (- \frac{1}{5 ( x ^{2} + 1 ) ^{5}}) : - \frac{1}{( x ^{2} + 1 ) ^{5}}

= - \frac{1}{( x ^{2} + 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{( x ^{2} + 1 ) ^{5}} + C

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What is the integral of (10x)/((x^2+1)^6) ?
The integral of (10x)/((x^2+1)^6) is -1/((x^2+1)^5)+C