What is the integral of x(6+x^2)^{10} ?

The integral of x(6+x^2)^{10} is 1/22 (6+x^2)^{11}+C

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integral of x(6+x^2)^{10}

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

\frac{1}{22} (6 + x^{2})^{11} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{10}}{2} d u

= \int \frac{u ^{10}}{2} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{11}}{11}

= \frac{1}{2} \cdot \frac{u ^{11}}{11}

Substitute back

u = 6 + x^{2}

= \frac{1}{2} \cdot \frac{( 6 + x ^{2} ) ^{11}}{11}

Simplify

\frac{1}{2} \cdot \frac{( 6 + x ^{2} ) ^{11}}{11} : \frac{1}{22} (6 + x^{2})^{11}

= \frac{1}{22} (6 + x^{2})^{11}

Add a constant to the solution

= \frac{1}{22} (6 + x^{2})^{11} + C

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What is the integral of x(6+x^2)^{10} ?
The integral of x(6+x^2)^{10} is 1/22 (6+x^2)^{11}+C