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

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

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

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

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

Solution steps

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

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 (y)

= sin (arctan (y))

Use the following identity:

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

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

Add a constant to the solution

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

f (x) = x ln (x + 1)

\frac{d}{d x} (sin (x) cos (x))

\frac{\partial}{\partial x} (\frac{7 x y}{x ^{2} + y ^{4}})

x \to 1 lim ((x - 1)^{(x - 1)})

\int_{0}^{x} 10 cos (x + y + z) d z

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