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

x^{2} \frac{d y}{d x} = \frac{1}{4} x^{2} + y^{2}

t \to 0 lim (e^{- 7 t} i + \frac{t ^{2}}{sin ^{2} ( t )} j + tan (3 t k))

f (x) = sin (x^{2} + 1)

\int 2 x^{3} \cdot e^{- x^{2}} d x

y^{^{''}} + 4 y = 8 cos (2 x)

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