What is the integral of x(9x^2+7)^3 ?

The integral of x(9x^2+7)^3 is 1/72 (9x^2+7)^4+C

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integral of x(9x^2+7)^3

\int x (9 x^{2} + 7)^{3} d x

\frac{1}{72} (9 x^{2} + 7)^{4} + C

Solution steps

\int x (9 x^{2} + 7)^{3} d x

Apply u-substitution

= \int \frac{u ^{3}}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u^{3} d u

Apply the Power Rule

= \frac{1}{18} \cdot \frac{u ^{4}}{4}

Substitute back

u = 9 x^{2} + 7

= \frac{1}{18} \cdot \frac{( 9 x ^{2} + 7 ) ^{4}}{4}

Simplify

\frac{1}{18} \cdot \frac{( 9 x ^{2} + 7 ) ^{4}}{4} : \frac{1}{72} (9 x^{2} + 7)^{4}

= \frac{1}{72} (9 x^{2} + 7)^{4}

Add a constant to the solution

= \frac{1}{72} (9 x^{2} + 7)^{4} + C

What is the integral of x(9x^2+7)^3 ?
The integral of x(9x^2+7)^3 is 1/72 (9x^2+7)^4+C