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

x \to 2 lim (\frac{x + 6}{x - 2})

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

\frac{d y}{d x} = - k y

x \to 3 lim (∣ x + 1 ∣ + ∣ x - 1 ∣)

arcsec (5 x)

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