What is the integral of x^3sqrt(x^2+39) ?

The integral of x^3sqrt(x^2+39) is 1/5 (x^2+39)^{5/2}-13(x^2+39)^{3/2}+C

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integral of x^3sqrt(x^2+39)

\int x^{3} x^{2} + 39 d x

\frac{1}{5} (x^{2} + 39)^{\frac{5}{2}} - 13 (x^{2} + 39)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 39 d x

Apply u-substitution

: \int (u^{2} - 39) u^{2} d u

= \int (u^{2} - 39) u^{2} d u

Expand

(u^{2} - 39) u^{2} : u^{4} - 39 u^{2}

= \int u^{4} - 39 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 39 u^{2} d u

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

\int 39 u^{2} d u = 13 u^{3}

= \frac{u ^{5}}{5} - 13 u^{3}

Substitute back

u = x^{2} + 39

= \frac{( x ^{2} + 39 ) ^{5}}{5} - 13 (x^{2} + 39)^{3}

Simplify

\frac{( x ^{2} + 39 ) ^{5}}{5} - 13 (x^{2} + 39)^{3} : \frac{1}{5} (x^{2} + 39)^{\frac{5}{2}} - 13 (x^{2} + 39)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 39)^{\frac{5}{2}} - 13 (x^{2} + 39)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 39)^{\frac{5}{2}} - 13 (x^{2} + 39)^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(x^2+39) ?
The integral of x^3sqrt(x^2+39) is 1/5 (x^2+39)^{5/2}-13(x^2+39)^{3/2}+C