What is the integral of 1/(xsqrt(2x-9)) ?

The integral of 1/(xsqrt(2x-9)) is 2/3 arctan(1/3 sqrt(2x-9))+C

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integral of 1/(xsqrt(2x-9))

\int \frac{1}{x 2 x - 9} d x

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

Solution steps

\int \frac{1}{x 2 x - 9} d x

Apply u-substitution

: \int \frac{2}{u ^{2} + 9} d u

= \int \frac{2}{u ^{2} + 9} d u

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 2 \cdot \frac{1}{3} arctan (\frac{2 x - 9}{3})

Simplify

2 \cdot \frac{1}{3} arctan (\frac{2 x - 9}{3}) : \frac{2}{3} arctan (\frac{1}{3} 2 x - 9)

= \frac{2}{3} arctan (\frac{1}{3} 2 x - 9)

Add a constant to the solution

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

\int \frac{9 x ^{3} - 9 x + 4}{x ^{3} - x ^{2}} d x

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\int (x) cos (x) d x

\int \frac{2 t + 1}{t ^{2} + 3 t - 1} d t

What is the integral of 1/(xsqrt(2x-9)) ?
The integral of 1/(xsqrt(2x-9)) is 2/3 arctan(1/3 sqrt(2x-9))+C