What is the integral of x^2(2x^3+1)^7 ?

The integral of x^2(2x^3+1)^7 is 1/48 (2x^3+1)^8+C

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integral of x^2(2x^3+1)^7

\int x^{2} (2 x^{3} + 1)^{7} d x

\frac{1}{48} (2 x^{3} + 1)^{8} + C

Solution steps

\int x^{2} (2 x^{3} + 1)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{6} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{1}{6} \cdot \frac{u ^{8}}{8}

Substitute back

u = 2 x^{3} + 1

= \frac{1}{6} \cdot \frac{( 2 x ^{3} + 1 ) ^{8}}{8}

Simplify

\frac{1}{6} \cdot \frac{( 2 x ^{3} + 1 ) ^{8}}{8} : \frac{1}{48} (2 x^{3} + 1)^{8}

= \frac{1}{48} (2 x^{3} + 1)^{8}

Add a constant to the solution

= \frac{1}{48} (2 x^{3} + 1)^{8} + C

e^{x} 5

\frac{d y}{d x} = (y^{2} + 1)

\int_{- 1}^{1} \frac{1}{x ^{2} - x - 2} d x

\int ln (1 + x^{4}) d x

n \to \infty lim (- 1)

What is the integral of x^2(2x^3+1)^7 ?
The integral of x^2(2x^3+1)^7 is 1/48 (2x^3+1)^8+C