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

The integral of 2x(1+x^2)^8 is 1/9 (1+x^2)^9+C

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

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

\frac{1}{9} (1 + x^{2})^{9} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x (1 + x^{2})^{8} d x

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{9}}{9}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{9}}{9}

Substitute back

u = 1 + x^{2}

= 2 \cdot \frac{1}{2} \cdot \frac{( 1 + x ^{2} ) ^{9}}{9}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( 1 + x ^{2} ) ^{9}}{9} : \frac{1}{9} (1 + x^{2})^{9}

= \frac{1}{9} (1 + x^{2})^{9}

Add a constant to the solution

= \frac{1}{9} (1 + x^{2})^{9} + C

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What is the integral of 2x(1+x^2)^8 ?
The integral of 2x(1+x^2)^8 is 1/9 (1+x^2)^9+C