What is the integral of x(x^2+6)^{67} ?

The integral of x(x^2+6)^{67} is 1/136 (x^2+6)^{68}+C

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integral of x(x^2+6)^{67}

\int x (x^{2} + 6)^{67} d x

\frac{1}{136} (x^{2} + 6)^{68} + C

Solution steps

\int x (x^{2} + 6)^{67} d x

Apply u-substitution

: \int \frac{u ^{67}}{2} d u

= \int \frac{u ^{67}}{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^{67} d u

Apply the Power Rule

: \frac{u ^{68}}{68}

= \frac{1}{2} \cdot \frac{u ^{68}}{68}

Substitute back

u = x^{2} + 6

= \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{68}}{68}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{68}}{68} : \frac{1}{136} (x^{2} + 6)^{68}

= \frac{1}{136} (x^{2} + 6)^{68}

Add a constant to the solution

= \frac{1}{136} (x^{2} + 6)^{68} + C

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What is the integral of x(x^2+6)^{67} ?
The integral of x(x^2+6)^{67} is 1/136 (x^2+6)^{68}+C