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

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

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

\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{d}{d x} (- (3 x + 2) cos (2 x))

f (x) = sin (x) + \frac{1}{2} cot (x)

n = 0 \sum \infty \frac{n}{2 n + 4}

\int - \frac{sin ( x ) tan ( x )}{1} d x

\frac{d y}{d t} - \frac{4 t}{( 1 + 4 t ^{2} )} y = 4 t

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