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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{3} \cdot \int (u + 9)^{\frac{1}{2}} d u

Apply u-substitution

: \int v^{\frac{1}{2}} d v

= \frac{1}{3} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{4.5 x - 4}{( 3 x - 4 ) ^{\frac{1}{2}}}

e^{x} + e^{x} x

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

\int_{0}^{1} (x^{3})^{2} d x

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