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

The integral of x^3sqrt(x^2+2) is 1/5 (x^2+2)^{5/2}-2/3 (x^2+2)^{3/2}+C

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

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

\frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int (u^{2} - 2) u^{2} d u

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

Expand

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

= \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

= \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}

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

Substitute back

u = x^{2} + 2

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

Simplify

\frac{( x ^{2} + 2 ) ^{5}}{5} - \frac{2 ( x ^{2} + 2 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}} + C

\frac{d}{d x} (- x^{- 1})

\int x^{2} cos (x y) d y

f (t) = e^{4 t}

\frac{d}{d x} (2 sin (x) + 2)

t \to 0 lim (\frac{6 e ^{t} - 6}{t})

What is the integral of x^3sqrt(x^2+2) ?
The integral of x^3sqrt(x^2+2) is 1/5 (x^2+2)^{5/2}-2/3 (x^2+2)^{3/2}+C