What is the integral of 1/((1+x^2)^{5/2)} ?

The integral of 1/((1+x^2)^{5/2)} is x/((1+x^2)^{1/2)}-(x^3)/(3(1+x^2)^{3/2)}+C

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integral of 1/((1+x^2)^{5/2)}

\int \frac{1}{( 1 + x ^{2} ) ^{\frac{5}{2}}} d x

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

Solution steps

\int \frac{1}{( 1 + x ^{2} ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int cos^{3} (u) d u

= \int cos^{3} (u) d u

Rewrite using trig identities

= \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v - \int v^{2} d v

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= v - \frac{v ^{3}}{3}

Substitute back

= sin (arctan (x)) - \frac{sin ^{3} ( arctan ( x ) )}{3}

Simplify

sin (arctan (x)) - \frac{sin ^{3} ( arctan ( x ) )}{3} : \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{x ^{3}}{3 ( 1 + x ^{2} ) ^{\frac{3}{2}}}

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

Add a constant to the solution

= \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{x ^{3}}{3 ( 1 + x ^{2} ) ^{\frac{3}{2}}} + C

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f (x) = 5^{- x}

\int \frac{2 x ^{3}}{2 x ^{4} + x ^{2} - 1} d x

x \to \infty lim (0 (x))

\int 15 x^{\frac{3}{2}} d x

What is the integral of 1/((1+x^2)^{5/2)} ?
The integral of 1/((1+x^2)^{5/2)} is x/((1+x^2)^{1/2)}-(x^3)/(3(1+x^2)^{3/2)}+C