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

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

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

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

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

Solution steps

\int (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

f (x) = x^{4} - 4 x^{3}

\int x \cdot 1 - x^{4} d x

n = 0 \sum \infty (\frac{1}{2})^{n + 1}

y^{^{'}} + \frac{y}{x + 1} = 0

\frac{d}{d x} (\frac{1}{9} (1 - (3 x + 1) e^{- 3 x}))

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