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

The integral of y^3sqrt(1+y) is 2/9 (1+y)^{9/2}-6/7 (1+y)^{7/2}+6/5 (1+y)^{5/2}-2/3 (1+y)^{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 y^3sqrt(1+y)

\int y^{3} 1 + y d y

\frac{2}{9} (1 + y)^{\frac{9}{2}} - \frac{6}{7} (1 + y)^{\frac{7}{2}} + \frac{6}{5} (1 + y)^{\frac{5}{2}} - \frac{2}{3} (1 + y)^{\frac{3}{2}} + C

Solution steps

\int y^{3} 1 + y d y

Apply u-substitution

: \int (u - 1)^{3} u d u

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

Expand

(u - 1)^{3} u : u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - u

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

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

\int 3 u^{\frac{5}{2}} d u = \frac{6}{7} u^{\frac{7}{2}}

\int 3 u^{\frac{3}{2}} d u = \frac{6}{5} u^{\frac{5}{2}}

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

= \frac{2}{9} u^{\frac{9}{2}} - \frac{6}{7} u^{\frac{7}{2}} + \frac{6}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + y

= \frac{2}{9} (1 + y)^{\frac{9}{2}} - \frac{6}{7} (1 + y)^{\frac{7}{2}} + \frac{6}{5} (1 + y)^{\frac{5}{2}} - \frac{2}{3} (1 + y)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (1 + y)^{\frac{9}{2}} - \frac{6}{7} (1 + y)^{\frac{7}{2}} + \frac{6}{5} (1 + y)^{\frac{5}{2}} - \frac{2}{3} (1 + y)^{\frac{3}{2}} + C

\int 2 tan (2 x) sec^{2} (2 x) d x

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

\int \frac{( 2 x + 1 )}{( x ^{2} + 9 )} d x

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

\int \frac{1}{5 - 2 x ^{2}} d x

What is the integral of y^3sqrt(1+y) ?
The integral of y^3sqrt(1+y) is 2/9 (1+y)^{9/2}-6/7 (1+y)^{7/2}+6/5 (1+y)^{5/2}-2/3 (1+y)^{3/2}+C