What is the integral of (x^3+x)(x^4+2x^2+9)^{1/2} ?

The integral of (x^3+x)(x^4+2x^2+9)^{1/2} is 1/6 (x^4+2x^2+9)^{3/2}+C

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integral of (x^3+x)(x^4+2x^2+9)^{1/2}

\int (x^{3} + x) (x^{4} + 2 x^{2} + 9)^{\frac{1}{2}} d x

\frac{1}{6} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}} + C

Solution steps

\int (x^{3} + x) (x^{4} + 2 x^{2} + 9)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{4} d u

= \int \frac{u ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{4} + 2 x^{2} + 9

= \frac{1}{4} \cdot \frac{2}{3} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}} : \frac{1}{6} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (x^{4} + 2 x^{2} + 9)^{\frac{3}{2}} + C

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What is the integral of (x^3+x)(x^4+2x^2+9)^{1/2} ?
The integral of (x^3+x)(x^4+2x^2+9)^{1/2} is 1/6 (x^4+2x^2+9)^{3/2}+C