What is the integral of 1/((x^2+1)^{3/2)} ?

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

Popular >

integral of 1/((x^2+1)^{3/2)}

integral

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= sin (u)

Substitute back

u = arctan (x)

= sin (arctan (x))

Use the following identity:

sin (arctan (x)) = \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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

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

sin^{5} (x) cos^{3} (x)

\frac{x ^{2}}{( x ^{2} - 1 ) ^{2}}

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

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