What is the integral of (81x^2)/(sqrt(9x-2)) ?

The integral of (81x^2)/(sqrt(9x-2)) is 2/9 (1/5 (9x-2)^{5/2}+4/3 (9x-2)^{3/2}+4sqrt(9x-2))+C

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integral of (81x^2)/(sqrt(9x-2))

integral

\frac{81 x ^{2}}{9 x - 2}

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

Solution steps

\int \frac{81 x ^{2}}{9 x - 2} d x

Take the constant out:

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

= 81 \cdot \int \frac{x ^{2}}{9 x - 2} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} + 2 ) ^{2}}{729} d u

= 81 \cdot \int \frac{2 ( u ^{2} + 2 ) ^{2}}{729} d u

Take the constant out:

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

= 81 \cdot \frac{2}{729} \cdot \int (u^{2} + 2)^{2} d u

Expand

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

= 81 \cdot \frac{2}{729} \cdot \int u^{4} + 4 u^{2} + 4 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

= 81 \cdot \frac{2}{729} (\int u^{4} d u + \int 4 u^{2} d u + \int 4 d u)

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

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

\int 4 d u = 4 u

= 81 \cdot \frac{2}{729} (\frac{u ^{5}}{5} + \frac{4 u ^{3}}{3} + 4 u)

Substitute back

u = 9 x - 2

= 81 \cdot \frac{2}{729} (\frac{( 9 x - 2 ) ^{5}}{5} + \frac{4 ( 9 x - 2 ) ^{3}}{3} + 4 9 x - 2)

Simplify

81 \cdot \frac{2}{729} (\frac{( 9 x - 2 ) ^{5}}{5} + \frac{4 ( 9 x - 2 ) ^{3}}{3} + 4 9 x - 2) : \frac{2}{9} (\frac{1}{5} (9 x - 2)^{\frac{5}{2}} + \frac{4}{3} (9 x - 2)^{\frac{3}{2}} + 4 9 x - 2)

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

Add a constant to the solution

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

4 cos^{2} (x) - 2 sin^{2} (x)

tan^{2} (3 x - 2)

2 sin^{2} (x)

\frac{3}{( x + 2 )}

\frac{ln ( x )}{( 1 + ln ( x ) ) ^{2}}

What is the integral of (81x^2)/(sqrt(9x-2)) ?
The integral of (81x^2)/(sqrt(9x-2)) is 2/9 (1/5 (9x-2)^{5/2}+4/3 (9x-2)^{3/2}+4sqrt(9x-2))+C