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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(x+1))

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

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

Solution steps

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

Apply u-substitution

: \int 2 (u^{2} - 1)^{3} d u

= \int 2 (u^{2} - 1)^{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (u^{2} - 1)^{3} d u

Expand

(u^{2} - 1)^{3} : u^{6} - 3 u^{4} + 3 u^{2} - 1

= 2 \cdot \int u^{6} - 3 u^{4} + 3 u^{2} - 1 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

= 2 (\int u^{6} d u - \int 3 u^{4} d u + \int 3 u^{2} d u - \int 1 d u)

\int u^{6} d u = \frac{u ^{7}}{7}

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

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

\int 1 d u = u

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( x ^{3} + 4 )}{x ( x ^{2} + 2 )} d x

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

\int (cos (3 t) - t^{2}) (sin (3 t) - t^{3})^{5} d t

\int \frac{t ^{2}}{a t ^{3} - b} d t

\int 70 csc^{2} (θ) d θ

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