What is the integral of x/(sqrt(9x+1)) ?

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

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integral of x/(sqrt(9x+1))

\int \frac{x}{9 x + 1} d x

\frac{2}{81} (\frac{1}{3} (9 x + 1)^{\frac{3}{2}} - 9 x + 1) + C

Solution steps

\int \frac{x}{9 x + 1} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} - 1 )}{81} d u

= \int \frac{2 ( u ^{2} - 1 )}{81} d u

Take the constant out:

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

= \frac{2}{81} \cdot \int 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

= \frac{2}{81} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= \frac{2}{81} (\frac{u ^{3}}{3} - u)

Substitute back

u = 9 x + 1

= \frac{2}{81} (\frac{( 9 x + 1 ) ^{3}}{3} - 9 x + 1)

Simplify

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

= \frac{2}{81} (\frac{1}{3} (9 x + 1)^{\frac{3}{2}} - 9 x + 1)

Add a constant to the solution

= \frac{2}{81} (\frac{1}{3} (9 x + 1)^{\frac{3}{2}} - 9 x + 1) + C

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