What is the integral of 9xsqrt(x^2+8) ?

The integral of 9xsqrt(x^2+8) is 3(x^2+8)^{3/2}+C

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integral of 9xsqrt(x^2+8)

\int 9 x x^{2} + 8 d x

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

Solution steps

\int 9 x x^{2} + 8 d x

Take the constant out:

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

= 9 \cdot \int x x^{2} + 8 d x

Apply u-substitution

= 9 \cdot \int u^{2} d u

Apply the Power Rule

= 9 \cdot \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 8

= 9 \cdot \frac{( x ^{2} + 8 ) ^{3}}{3}

Simplify

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

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

Add a constant to the solution

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

What is the integral of 9xsqrt(x^2+8) ?
The integral of 9xsqrt(x^2+8) is 3(x^2+8)^{3/2}+C