What is the integral of (x^2)/(sqrt(x-4)) ?

The integral of (x^2)/(sqrt(x-4)) is 2(1/5 (x-4)^{5/2}+8/3 (x-4)^{3/2}+16sqrt(x-4))+C

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integral of (x^2)/(sqrt(x-4))

\int \frac{x ^{2}}{x - 4} d x

2 (\frac{1}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}} + 16 x - 4) + C

Solution steps

\int \frac{x ^{2}}{x - 4} d x

Apply u-substitution

: \int 2 (u^{2} + 4)^{2} d u

= \int 2 (u^{2} + 4)^{2} 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} + 4)^{2} d u

Expand

(u^{2} + 4)^{2} : u^{4} + 8 u^{2} + 16

= 2 \cdot \int u^{4} + 8 u^{2} + 16 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^{4} d u + \int 8 u^{2} d u + \int 16 d u)

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

\int 8 u^{2} d u = \frac{8 u ^{3}}{3}

\int 16 d u = 16 u

= 2 (\frac{u ^{5}}{5} + \frac{8 u ^{3}}{3} + 16 u)

Substitute back

u = x - 4

= 2 (\frac{( x - 4 ) ^{5}}{5} + \frac{8 ( x - 4 ) ^{3}}{3} + 16 x - 4)

Simplify

2 (\frac{( x - 4 ) ^{5}}{5} + \frac{8 ( x - 4 ) ^{3}}{3} + 16 x - 4) : 2 (\frac{1}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}} + 16 x - 4)

= 2 (\frac{1}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}} + 16 x - 4)

Add a constant to the solution

= 2 (\frac{1}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}} + 16 x - 4) + C

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What is the integral of (x^2)/(sqrt(x-4)) ?
The integral of (x^2)/(sqrt(x-4)) is 2(1/5 (x-4)^{5/2}+8/3 (x-4)^{3/2}+16sqrt(x-4))+C