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

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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 e^{- 0.6 x} d x

y = 9 x^{2} + 3 x - 9

- \frac{y ^{2}}{t ^{2}} d t + \frac{2 y}{t} d y = 0

x \to - 3 lim (- h (x))

(5 x^{6} + 4 x^{3})^{4}

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