What is the integral of x^3sqrt(x^2+47) ?

The integral of x^3sqrt(x^2+47) is 1/5 (x^2+47)^{5/2}-47/3 (x^2+47)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^3sqrt(x^2+47)

\int x^{3} x^{2} + 47 d x

\frac{1}{5} (x^{2} + 47)^{\frac{5}{2}} - \frac{47}{3} (x^{2} + 47)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 47 d x

Apply u-substitution

: \int (u^{2} - 47) u^{2} d u

= \int (u^{2} - 47) u^{2} d u

Expand

(u^{2} - 47) u^{2} : u^{4} - 47 u^{2}

= \int u^{4} - 47 u^{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

= \int u^{4} d u - \int 47 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 47 u^{2} d u = \frac{47 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{47 u ^{3}}{3}

Substitute back

u = x^{2} + 47

= \frac{( x ^{2} + 47 ) ^{5}}{5} - \frac{47 ( x ^{2} + 47 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 47 ) ^{5}}{5} - \frac{47 ( x ^{2} + 47 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 47)^{\frac{5}{2}} - \frac{47}{3} (x^{2} + 47)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 47)^{\frac{5}{2}} - \frac{47}{3} (x^{2} + 47)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 47)^{\frac{5}{2}} - \frac{47}{3} (x^{2} + 47)^{\frac{3}{2}} + C

\int - 3 e^{3 x} d x

\int x^{3} x^{2} + 48 d x

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

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

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

What is the integral of x^3sqrt(x^2+47) ?
The integral of x^3sqrt(x^2+47) is 1/5 (x^2+47)^{5/2}-47/3 (x^2+47)^{3/2}+C