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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

= \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

\int 4 x^{3} + \frac{7}{x} d x

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

\int \frac{3}{2 x + 1} d x

\int \frac{1}{4 6 - 9 x} d x

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

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