What is the integral of 10x(5x^2+9)^7 ?

The integral of 10x(5x^2+9)^7 is 1/8 (5x^2+9)^8+C

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integral of 10x(5x^2+9)^7

\int 10 x (5 x^{2} + 9)^{7} d x

\frac{1}{8} (5 x^{2} + 9)^{8} + C

Solution steps

\int 10 x (5 x^{2} + 9)^{7} d x

Take the constant out:

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

= 10 \cdot \int x (5 x^{2} + 9)^{7} d x

Apply u-substitution

= 10 \cdot \int \frac{u ^{7}}{10} d u

Take the constant out:

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

= 10 \cdot \frac{1}{10} \cdot \int u^{7} d u

Apply the Power Rule

= 10 \cdot \frac{1}{10} \cdot \frac{u ^{8}}{8}

Substitute back

u = 5 x^{2} + 9

= 10 \cdot \frac{1}{10} \cdot \frac{( 5 x ^{2} + 9 ) ^{8}}{8}

Simplify

10 \cdot \frac{1}{10} \cdot \frac{( 5 x ^{2} + 9 ) ^{8}}{8} : \frac{1}{8} (5 x^{2} + 9)^{8}

= \frac{1}{8} (5 x^{2} + 9)^{8}

Add a constant to the solution

= \frac{1}{8} (5 x^{2} + 9)^{8} + C

What is the integral of 10x(5x^2+9)^7 ?
The integral of 10x(5x^2+9)^7 is 1/8 (5x^2+9)^8+C