What is the integral of ((x^3+2x))/(sqrt(x^2+4)) ?

The integral of ((x^3+2x))/(sqrt(x^2+4)) is 1/3 (x^2+4)^{3/2}-2sqrt(x^2+4)+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+2x))/(sqrt(x^2+4))

\int \frac{( x ^{3} + 2 x )}{x ^{2} + 4} d x

\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4 + C

Solution steps

\int \frac{x ^{3} + 2 x}{x ^{2} + 4} d x

Expand

\frac{x ^{3} + 2 x}{x ^{2} + 4} : \frac{x ^{3}}{x ^{2} + 4} + \frac{2 x}{x ^{2} + 4}

= \int \frac{x ^{3}}{x ^{2} + 4} + \frac{2 x}{x ^{2} + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{3}}{x ^{2} + 4} d x + \int \frac{2 x}{x ^{2} + 4} d x

\int \frac{x ^{3}}{x ^{2} + 4} d x = \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4

\int \frac{2 x}{x ^{2} + 4} d x = 2 x^{2} + 4

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4 + 2 x^{2} + 4

Simplify

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4

Add a constant to the solution

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 2 x^{2} + 4 + C

\int 2^{x} \cdot 2^{2 x} d x

\int 12 - 3 t^{2} d t

\int 5 t^{3} d t

\int \frac{5 x ^{2} - 5}{x ^{3} - 9 x ^{2} + 23 x - 15} d x

\int sin^{8} (\frac{3}{4}) x \cdot cos^{6} (\frac{3}{4}) x d x

What is the integral of ((x^3+2x))/(sqrt(x^2+4)) ?
The integral of ((x^3+2x))/(sqrt(x^2+4)) is 1/3 (x^2+4)^{3/2}-2sqrt(x^2+4)+C