What is the integral of (x^2)/(sqrt(2x-3)) ?

The integral of (x^2)/(sqrt(2x-3)) is 1/5 (x^2+2x+6)sqrt(2x-3)+C

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integral of (x^2)/(sqrt(2x-3))

\int \frac{x ^{2}}{2 x - 3} d x

\frac{1}{5} (x^{2} + 2 x + 6) 2 x - 3 + C

Solution steps

\int \frac{x ^{2}}{2 x - 3} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{4} \cdot \int (u^{2} + 3)^{2} d u

Expand

(u^{2} + 3)^{2} : u^{4} + 6 u^{2} + 9

= \frac{1}{4} \cdot \int u^{4} + 6 u^{2} + 9 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

= \frac{1}{4} (\int u^{4} d u + \int 6 u^{2} d u + \int 9 d u)

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

\int 6 u^{2} d u = 2 u^{3}

\int 9 d u = 9 u

= \frac{1}{4} (\frac{u ^{5}}{5} + 2 u^{3} + 9 u)

Substitute back

u = 2 x - 3

= \frac{1}{4} (\frac{( 2 x - 3 ) ^{5}}{5} + 2 (2 x - 3)^{3} + 9 2 x - 3)

Simplify

\frac{1}{4} (\frac{( 2 x - 3 ) ^{5}}{5} + 2 (2 x - 3)^{3} + 9 2 x - 3) : \frac{1}{5} (x^{2} + 2 x + 6) 2 x - 3

= \frac{1}{5} (x^{2} + 2 x + 6) 2 x - 3

Add a constant to the solution

= \frac{1}{5} (x^{2} + 2 x + 6) 2 x - 3 + C

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What is the integral of (x^2)/(sqrt(2x-3)) ?
The integral of (x^2)/(sqrt(2x-3)) is 1/5 (x^2+2x+6)sqrt(2x-3)+C