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

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

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integral of (x^{3/2}+7)^6sqrt(x)

\int (x^{\frac{3}{2}} + 7)^{6} x d x

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

Solution steps

\int (x^{\frac{3}{2}} + 7)^{6} x d x

Apply u-substitution

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

Substitute back

u = x^{\frac{3}{2}} + 7

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

Simplify

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

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

Add a constant to the solution

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

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