What is the integral of 3x^3sqrt(x^2+2) ?

The integral of 3x^3sqrt(x^2+2) is 3/5 (x^2+2)^{5/2}-2(x^2+2)^{3/2}+C

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integral of 3x^3sqrt(x^2+2)

\int 3 x^{3} x^{2} + 2 d x

\frac{3}{5} (x^{2} + 2)^{\frac{5}{2}} - 2 (x^{2} + 2)^{\frac{3}{2}} + C

Solution steps

\int 3 x^{3} x^{2} + 2 d x

Take the constant out:

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

= 3 \cdot \int x^{3} x^{2} + 2 d x

Apply u-substitution

= 3 \cdot \int (u^{2} - 2) u^{2} d u

Expand

(u^{2} - 2) u^{2} : u^{4} - 2 u^{2}

= 3 \cdot \int u^{4} - 2 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int u^{4} d u - \int 2 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

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

= 3 (\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3})

Substitute back

u = x^{2} + 2

= 3 (\frac{( x ^{2} + 2 ) ^{5}}{5} - \frac{2 ( x ^{2} + 2 ) ^{3}}{3})

Simplify

3 (\frac{( x ^{2} + 2 ) ^{5}}{5} - \frac{2 ( x ^{2} + 2 ) ^{3}}{3}) : \frac{3}{5} (x^{2} + 2)^{\frac{5}{2}} - 2 (x^{2} + 2)^{\frac{3}{2}}

= \frac{3}{5} (x^{2} + 2)^{\frac{5}{2}} - 2 (x^{2} + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{3}{5} (x^{2} + 2)^{\frac{5}{2}} - 2 (x^{2} + 2)^{\frac{3}{2}} + C

What is the integral of 3x^3sqrt(x^2+2) ?
The integral of 3x^3sqrt(x^2+2) is 3/5 (x^2+2)^{5/2}-2(x^2+2)^{3/2}+C