What is the integral of (7x^6+9)sqrt(x^7+9x) ?

The integral of (7x^6+9)sqrt(x^7+9x) is 2/3 (x^7+9x)^{3/2}+C

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integral of (7x^6+9)sqrt(x^7+9x)

\int (7 x^{6} + 9) x^{7} + 9 x d x

\frac{2}{3} (x^{7} + 9 x)^{\frac{3}{2}} + C

Solution steps

\int (7 x^{6} + 9) x^{7} + 9 x d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = x^{7} + 9 x

= \frac{2}{3} (x^{7} + 9 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x^{7} + 9 x)^{\frac{3}{2}} + C

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What is the integral of (7x^6+9)sqrt(x^7+9x) ?
The integral of (7x^6+9)sqrt(x^7+9x) is 2/3 (x^7+9x)^{3/2}+C