What is the integral of (x^3)/(sqrt(x^2+25)) ?

The integral of (x^3)/(sqrt(x^2+25)) is 1/3 (x^2+25)^{3/2}-25sqrt(x^2+25)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(x^2+25))

\int \frac{x ^{3}}{x ^{2} + 25} d x

\frac{1}{3} (x^{2} + 25)^{\frac{3}{2}} - 25 x^{2} + 25 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 25} d x

Apply u-substitution

: \int u^{2} - 25 d u

= \int u^{2} - 25 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 25 d u

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

\int 25 d u = 25 u

= \frac{u ^{3}}{3} - 25 u

Substitute back

u = x^{2} + 25

= \frac{( x ^{2} + 25 ) ^{3}}{3} - 25 x^{2} + 25

Simplify

\frac{( x ^{2} + 25 ) ^{3}}{3} - 25 x^{2} + 25 : \frac{1}{3} (x^{2} + 25)^{\frac{3}{2}} - 25 x^{2} + 25

= \frac{1}{3} (x^{2} + 25)^{\frac{3}{2}} - 25 x^{2} + 25

Add a constant to the solution

= \frac{1}{3} (x^{2} + 25)^{\frac{3}{2}} - 25 x^{2} + 25 + C

\int \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (3 x^{2} y + 4 x y^{2})

x \frac{d y}{d x} - 2 y = 2

\frac{\partial}{\partial y} (x ln (x^{2} + y^{2}))

\frac{d}{d x} (400 (15 - x^{2}) (3 x - 2))

What is the integral of (x^3)/(sqrt(x^2+25)) ?
The integral of (x^3)/(sqrt(x^2+25)) is 1/3 (x^2+25)^{3/2}-25sqrt(x^2+25)+C