What is the integral of (x+1)^2sqrt(3+x) ?

The integral of (x+1)^2sqrt(3+x) is 2/7 (3+x)^{7/2}-8/5 (3+x)^{5/2}+8/3 (3+x)^{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+1)^2sqrt(3+x)

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

\frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{8}{5} (3 + x)^{\frac{5}{2}} + \frac{8}{3} (3 + x)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Expand

u (u - 2)^{2} : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

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

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

\int 4 u^{\frac{3}{2}} d u = \frac{8}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{8}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}}

Substitute back

u = 3 + x

= \frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{8}{5} (3 + x)^{\frac{5}{2}} + \frac{8}{3} (3 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (3 + x)^{\frac{7}{2}} - \frac{8}{5} (3 + x)^{\frac{5}{2}} + \frac{8}{3} (3 + x)^{\frac{3}{2}} + C

\int - 1.09 + 5.8 e^{- 0.25 x} d x

\int \frac{1}{e ^{x} + 9 e ^{- x}} d x

\int 16 - 3 x^{2} - 2 x d x

\int x cos (y x) d x

\int \frac{sin ^{2} ( x ) - k}{sin ^{2} ( x )} d x

What is the integral of (x+1)^2sqrt(3+x) ?
The integral of (x+1)^2sqrt(3+x) is 2/7 (3+x)^{7/2}-8/5 (3+x)^{5/2}+8/3 (3+x)^{3/2}+C