What is the integral of x(x^2+9)^{1/2} ?

The integral of x(x^2+9)^{1/2} is 1/3 x^2(x^2+9)^{1/2}+3(x^2+9)^{1/2}+C

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integral of x(x^2+9)^{1/2}

\int x (x^{2} + 9)^{\frac{1}{2}} d x

\frac{1}{3} x^{2} (x^{2} + 9)^{\frac{1}{2}} + 3 (x^{2} + 9)^{\frac{1}{2}} + C

Solution steps

\int x (x^{2} + 9)^{\frac{1}{2}} d x

Apply Integration By Parts

: \frac{1}{2} x^{2} (x^{2} + 9)^{\frac{1}{2}} - \int \frac{x ^{3}}{2 ( x ^{2} + 9 ) ^{\frac{1}{2}}} d x

= \frac{1}{2} x^{2} (x^{2} + 9)^{\frac{1}{2}} - \int \frac{x ^{3}}{2 ( x ^{2} + 9 ) ^{\frac{1}{2}}} d x

\int \frac{x ^{3}}{2 ( x ^{2} + 9 ) ^{\frac{1}{2}}} d x = \frac{1}{6} (x^{2} - 18) (x^{2} + 9)^{\frac{1}{2}}

= \frac{1}{2} x^{2} (x^{2} + 9)^{\frac{1}{2}} - \frac{1}{6} (x^{2} - 18) (x^{2} + 9)^{\frac{1}{2}}

Simplify

\frac{1}{2} x^{2} (x^{2} + 9)^{\frac{1}{2}} - \frac{1}{6} (x^{2} - 18) (x^{2} + 9)^{\frac{1}{2}} : \frac{1}{3} x^{2} (x^{2} + 9)^{\frac{1}{2}} + 3 (x^{2} + 9)^{\frac{1}{2}}

= \frac{1}{3} x^{2} (x^{2} + 9)^{\frac{1}{2}} + 3 (x^{2} + 9)^{\frac{1}{2}}

Add a constant to the solution

= \frac{1}{3} x^{2} (x^{2} + 9)^{\frac{1}{2}} + 3 (x^{2} + 9)^{\frac{1}{2}} + C

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What is the integral of x(x^2+9)^{1/2} ?
The integral of x(x^2+9)^{1/2} is 1/3 x^2(x^2+9)^{1/2}+3(x^2+9)^{1/2}+C