What is the integral of 7x^2(-2x^3-5)^6 ?

The integral of 7x^2(-2x^3-5)^6 is 1/6 (2x^3+5)^7+C

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integral of 7x^2(-2x^3-5)^6

\int 7 x^{2} (- 2 x^{3} - 5)^{6} d x

\frac{1}{6} (2 x^{3} + 5)^{7} + C

Solution steps

\int 7 x^{2} (- 2 x^{3} - 5)^{6} d x

Take the constant out:

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

= 7 \cdot \int x^{2} (- 2 x^{3} - 5)^{6} d x

Factor

- 2 x^{3} - 5 : - (2 x^{3} + 5)

= 7 \cdot \int x^{2} (- (2 x^{3} + 5))^{6} d x

(- (2 x^{3} + 5))^{6} = (2 x^{3} + 5)^{6}

= 7 \cdot \int x^{2} (2 x^{3} + 5)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{6} d u

= 7 \cdot \int \frac{u ^{6}}{6} d u

Take the constant out:

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

= 7 \cdot \frac{1}{6} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 7 \cdot \frac{1}{6} \cdot \frac{u ^{7}}{7}

Substitute back

u = 2 x^{3} + 5

= 7 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 5 ) ^{7}}{7}

Simplify

7 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 5 ) ^{7}}{7} : \frac{1}{6} (2 x^{3} + 5)^{7}

= \frac{1}{6} (2 x^{3} + 5)^{7}

Add a constant to the solution

= \frac{1}{6} (2 x^{3} + 5)^{7} + C

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What is the integral of 7x^2(-2x^3-5)^6 ?
The integral of 7x^2(-2x^3-5)^6 is 1/6 (2x^3+5)^7+C