What is the integral of (x^{3/2}+8)^6sqrt(x) ?

The integral of (x^{3/2}+8)^6sqrt(x) is 2/21 (x^{3/2}+8)^7+C

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integral of (x^{3/2}+8)^6sqrt(x)

\int (x^{\frac{3}{2}} + 8)^{6} x d x

\frac{2}{21} (x^{\frac{3}{2}} + 8)^{7} + C

Solution steps

\int (x^{\frac{3}{2}} + 8)^{6} x d x

Apply u-substitution

: \int \frac{2 u ^{6}}{3} d u

= \int \frac{2 u ^{6}}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int u^{6} d u

Apply the Power Rule

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

= \frac{2}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = x^{\frac{3}{2}} + 8

= \frac{2}{3} \cdot \frac{( x ^{\frac{3}{2}} + 8 ) ^{7}}{7}

Simplify

\frac{2}{3} \cdot \frac{( x ^{\frac{3}{2}} + 8 ) ^{7}}{7} : \frac{2}{21} (x^{\frac{3}{2}} + 8)^{7}

= \frac{2}{21} (x^{\frac{3}{2}} + 8)^{7}

Add a constant to the solution

= \frac{2}{21} (x^{\frac{3}{2}} + 8)^{7} + C

n = 1 \sum \infty (- \frac{10}{11})^{n}

\frac{d}{d x} (\frac{x}{y ^{2}} - \frac{y}{x ^{2}})

f (x) = \frac{1}{1 + x ^{2}}, at x = - 1

\frac{\partial}{\partial θ} (n ln (2 θ))

\int \frac{50 x}{sec ( 5 x ^{2} + 5 )} d x

What is the integral of (x^{3/2}+8)^6sqrt(x) ?
The integral of (x^{3/2}+8)^6sqrt(x) is 2/21 (x^{3/2}+8)^7+C