What is the integral of 1/((x^2+4)^{1.5)} ?

The integral of 1/((x^2+4)^{1.5)} is x/((x^2+4)sqrt(x^2+4))+(x^3)/(4(4+x^2)^{3/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((x^2+4)^{1.5)}

\int \frac{1}{( x ^{2} + 4 ) ^{1.5}} d x

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

Solution steps

\int \frac{1}{( x ^{2} + 4 ) ^{1.5}} d x

(x^{2} + 4)^{1.5} = (x^{2} + 4) x^{2} + 4

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

Apply Integration By Parts

= \frac{x}{( x ^{2} + 4 ) x ^{2} + 4} - \int - \frac{3 x ^{2}}{( x ^{2} + 4 ) ^{\frac{5}{2}}} d x

\int - \frac{3 x ^{2}}{( x ^{2} + 4 ) ^{\frac{5}{2}}} d x = - \frac{x ^{3}}{4 ( 4 + x ^{2} ) ^{\frac{3}{2}}}

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

Simplify

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

Add a constant to the solution

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

What is the integral of 1/((x^2+4)^{1.5)} ?
The integral of 1/((x^2+4)^{1.5)} is x/((x^2+4)sqrt(x^2+4))+(x^3)/(4(4+x^2)^{3/2)}+C