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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int (5 x + 3) (x^{2} - 9) d x

\frac{d}{d x} (x^{2} + y^{2} + x y)

y = 2 x - 2 x^{2}, [0, 1]

y = \frac{1}{x ^{2}}, y = 0, x = 1, x = 5

x \to 0 lim (\frac{sin ( 2 x )}{x} + 4)

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