What is the integral of x^2*sqrt(1+x) ?

The integral of x^2*sqrt(1+x) is 2/7 (1+x)^{7/2}-4/5 (1+x)^{5/2}+2/3 (1+x)^{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^2*sqrt(1+x)

\int x^{2} \cdot 1 + x d x

\frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 1 + x d x

Apply u-substitution

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

Expand

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

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + x

= \frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (1 + x)^{\frac{7}{2}} - \frac{4}{5} (1 + x)^{\frac{5}{2}} + \frac{2}{3} (1 + x)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (- e^{8 y} cos (8 x) \cdot 8)

d er i v a t i v e y = tan (π x)

\frac{d}{d x} (2 x e^{2 y} - x cos (x y) + 2 y)

x \to - 2 lim (6 - 3 x - 3)

\int \frac{x ^{3}}{x ln ( x )} d x

What is the integral of x^2*sqrt(1+x) ?
The integral of x^2*sqrt(1+x) is 2/7 (1+x)^{7/2}-4/5 (1+x)^{5/2}+2/3 (1+x)^{3/2}+C