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

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

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

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

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

Solution steps

\int (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

\int_{- \infty}^{3} \frac{3}{x ^{2} + 9} d x

f (x) = x + 2, x = 1, x = 3

y^{^{'}} = \frac{x - y}{x + 2 y}

\frac{d ^{2}}{d x ^{2}} (\frac{x}{4 - 6 x})

t^{2} + 2 t + 50

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