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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3)/((x^2+1)^{5/2)}

\int \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{5}{2}}} d x

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

Solution steps

\int \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{5}{2}}} d x

Apply u-substitution

= \int \frac{u - 1}{2 u ^{\frac{5}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 1}{u ^{\frac{5}{2}}} d u

Expand

\frac{u - 1}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{3}{2}}} - \frac{1}{u ^{\frac{5}{2}}}

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{3}{2}}} - \frac{1}{u ^{\frac{5}{2}}} d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{1}{u ^{\frac{3}{2}}} d u - \int \frac{1}{u ^{\frac{5}{2}}} d u)

\int \frac{1}{u ^{\frac{3}{2}}} d u = - \frac{2}{u ^{\frac{1}{2}}}

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{3 ^{x + 1} - 6 ^{x - 1}}{1 2 ^{x}} d x

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

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

\int (\frac{6 x ^{2} + 2 x}{2 x}) d x

\int 3^{2 x + 4} + ln (3 - x) - e^{2 - 3 x} d x

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