What is the integral of (x^3)/(sqrt(x^2+121)) ?

The integral of (x^3)/(sqrt(x^2+121)) is 1/3 (x^2+121)^{3/2}-121sqrt(x^2+121)+C

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integral of (x^3)/(sqrt(x^2+121))

\int \frac{x ^{3}}{x ^{2} + 121} d x

\frac{1}{3} (x^{2} + 121)^{\frac{3}{2}} - 121 x^{2} + 121 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 121} d x

Apply u-substitution

: \int u^{2} - 121 d u

= \int u^{2} - 121 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 121 d u

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

\int 121 d u = 121 u

= \frac{u ^{3}}{3} - 121 u

Substitute back

u = x^{2} + 121

= \frac{( x ^{2} + 121 ) ^{3}}{3} - 121 x^{2} + 121

Simplify

\frac{( x ^{2} + 121 ) ^{3}}{3} - 121 x^{2} + 121 : \frac{1}{3} (x^{2} + 121)^{\frac{3}{2}} - 121 x^{2} + 121

= \frac{1}{3} (x^{2} + 121)^{\frac{3}{2}} - 121 x^{2} + 121

Add a constant to the solution

= \frac{1}{3} (x^{2} + 121)^{\frac{3}{2}} - 121 x^{2} + 121 + C

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

\int sin (22 x) sin (23 x) d x

\int \frac{- 1}{t} d t

x \to 1 + lim (x^{\frac{x}{1 - x}})

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 4

What is the integral of (x^3)/(sqrt(x^2+121)) ?
The integral of (x^3)/(sqrt(x^2+121)) is 1/3 (x^2+121)^{3/2}-121sqrt(x^2+121)+C