What is the integral of x^3sqrt(x^2+19) ?

The integral of x^3sqrt(x^2+19) is 1/5 (x^2+19)^{5/2}-19/3 (x^2+19)^{3/2}+C

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integral of x^3sqrt(x^2+19)

\int x^{3} x^{2} + 19 d x

\frac{1}{5} (x^{2} + 19)^{\frac{5}{2}} - \frac{19}{3} (x^{2} + 19)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 19 d x

Apply u-substitution

: \int (u^{2} - 19) u^{2} d u

= \int (u^{2} - 19) u^{2} d u

Expand

(u^{2} - 19) u^{2} : u^{4} - 19 u^{2}

= \int u^{4} - 19 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 19 u^{2} d u

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

\int 19 u^{2} d u = \frac{19 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{19 u ^{3}}{3}

Substitute back

u = x^{2} + 19

= \frac{( x ^{2} + 19 ) ^{5}}{5} - \frac{19 ( x ^{2} + 19 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 19 ) ^{5}}{5} - \frac{19 ( x ^{2} + 19 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 19)^{\frac{5}{2}} - \frac{19}{3} (x^{2} + 19)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 19)^{\frac{5}{2}} - \frac{19}{3} (x^{2} + 19)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 19)^{\frac{5}{2}} - \frac{19}{3} (x^{2} + 19)^{\frac{3}{2}} + C

y = \frac{4}{5 x + 1}, (- 1, - 1)

\frac{9}{x ^{2}} - \frac{3}{x}

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

\int_{- \infty}^{6} \frac{1}{x ^{2} + 1} d x

\frac{d}{d x} (2 e^{- x^{3}})

What is the integral of x^3sqrt(x^2+19) ?
The integral of x^3sqrt(x^2+19) is 1/5 (x^2+19)^{5/2}-19/3 (x^2+19)^{3/2}+C