What is the integral of (2x)/((x^2+1)^{90)} ?

The integral of (2x)/((x^2+1)^{90)} is -1/(89(x^2+1)^{89)}+C

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integral of (2x)/((x^2+1)^{90)}

\int \frac{2 x}{( x ^{2} + 1 ) ^{90}} d x

- \frac{1}{89 ( x ^{2} + 1 ) ^{89}} + C

Solution steps

\int \frac{2 x}{( x ^{2} + 1 ) ^{90}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{90}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{90}} d u

= 2 \cdot \int \frac{1}{2 u ^{90}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{90}} d u

Apply the Power Rule

: - \frac{1}{89 u ^{89}}

= 2 \cdot \frac{1}{2} (- \frac{1}{89 u ^{89}})

Substitute back

u = x^{2} + 1

= 2 \cdot \frac{1}{2} (- \frac{1}{89 ( x ^{2} + 1 ) ^{89}})

Simplify

2 \cdot \frac{1}{2} (- \frac{1}{89 ( x ^{2} + 1 ) ^{89}}) : - \frac{1}{89 ( x ^{2} + 1 ) ^{89}}

= - \frac{1}{89 ( x ^{2} + 1 ) ^{89}}

Add a constant to the solution

= - \frac{1}{89 ( x ^{2} + 1 ) ^{89}} + C

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What is the integral of (2x)/((x^2+1)^{90)} ?
The integral of (2x)/((x^2+1)^{90)} is -1/(89(x^2+1)^{89)}+C