What is the integral of (x^3)/(sqrt(x^2+36)) ?

The integral of (x^3)/(sqrt(x^2+36)) is 1/3 (x^2+36)^{3/2}-36sqrt(x^2+36)+C

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integral of (x^3)/(sqrt(x^2+36))

\int \frac{x ^{3}}{x ^{2} + 36} d x

\frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 36} d x

Apply u-substitution

: \int u^{2} - 36 d u

= \int u^{2} - 36 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^{2} d u - \int 36 d u

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

\int 36 d u = 36 u

= \frac{u ^{3}}{3} - 36 u

Substitute back

u = x^{2} + 36

= \frac{( x ^{2} + 36 ) ^{3}}{3} - 36 x^{2} + 36

Simplify

\frac{( x ^{2} + 36 ) ^{3}}{3} - 36 x^{2} + 36 : \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36

= \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36

Add a constant to the solution

= \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36 + C

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What is the integral of (x^3)/(sqrt(x^2+36)) ?
The integral of (x^3)/(sqrt(x^2+36)) is 1/3 (x^2+36)^{3/2}-36sqrt(x^2+36)+C