What is the integral of (x^3)/(sqrt(x^2+5)) ?

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

\int \frac{x ^{3}}{x ^{2} + 5} d x

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

Solution steps

\int \frac{x ^{3}}{x ^{2} + 5} d x

Apply u-substitution

: \int u^{2} - 5 d u

= \int u^{2} - 5 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 5 d u

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

\int 5 d u = 5 u

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

Substitute back

u = x^{2} + 5

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

Simplify

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

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

Add a constant to the solution

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

\int (2 x + 4) d x

\frac{d}{d x} (tan (\frac{π}{3}))

\int - \frac{6 x ^{3} + 8 x}{x ^{4} - 16} d x

\frac{d y}{d t} = 0.1 ln (\frac{4000}{y}) y

ln (7 x)

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