What is the integral of (x+1)\sqrt[3]{4-x} ?

The integral of (x+1)\sqrt[3]{4-x} is 3/7 (4-x)^{7/3}-15/4 (4-x)^{4/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x+1)\sqrt[3]{4-x}

\int (x + 1) 3 4 - x d x

\frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C

Solution steps

\int (x + 1) 3 4 - x d x

Expand

(x + 1) 3 4 - x : x 3 4 - x + 3 4 - x

= \int x 3 4 - x + 3 4 - x d x

Apply the Sum Rule:

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

= \int x 3 4 - x d x + \int 3 4 - x d x

\int x 3 4 - x d x = \frac{3}{7} (4 - x)^{\frac{7}{3}} - 3 (4 - x)^{\frac{4}{3}}

\int 3 4 - x d x = - \frac{3}{4} (4 - x)^{\frac{4}{3}}

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - 3 (4 - x)^{\frac{4}{3}} - \frac{3}{4} (4 - x)^{\frac{4}{3}}

Add similar elements:

- 3 (- x + 4)^{\frac{4}{3}} - \frac{3}{4} (- x + 4)^{\frac{4}{3}} = - \frac{15}{4} (- x + 4)^{\frac{4}{3}}

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C

\int (y^{2})^{2} d y

\int \frac{e ^{4 x}}{e ^{4 x} + 7} d x

\int x + x^{6} + x^{- 6} d x

\int (cos^{2} (x)) sin (x) d x

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

What is the integral of (x+1)\sqrt[3]{4-x} ?
The integral of (x+1)\sqrt[3]{4-x} is 3/7 (4-x)^{7/3}-15/4 (4-x)^{4/3}+C