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

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

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integral of 6x^2(2x^3+5)^5

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ( 2 u ^{\frac{3}{2}} + 5 ) ^{5}}{2 x} d u

= 6 \cdot \int \frac{( 2 u + 5 ) ^{5}}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int (2 u + 5)^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{2} d v

= 6 \cdot \frac{1}{3} \cdot \int \frac{v ^{5}}{2} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{6}}{6}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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