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

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

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

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

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

Solution steps

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

\frac{\partial}{\partial y} (T x^{5} y^{4} z^{8})

\frac{x ^{2}}{x ^{3} - 3 x}, \frac{1}{x ^{3} - 3 x}, 2, 11

\frac{d}{d x} (\frac{x e ^{x}}{sin ( x ) + 1})

tan^{2} (3 x)

y = 6 x^{3} ln (5 x)

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