What is the integral of 1/((x+3)sqrt(x-2)) ?

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

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integral of 1/((x+3)sqrt(x-2))

\int \frac{1}{( x + 3 ) x - 2} d x

\frac{2}{5} arctan (\frac{1}{5} (x - 2)) + C

Solution steps

\int \frac{1}{( x + 3 ) x - 2} d x

Apply u-substitution

: \int \frac{2}{u ^{2} + 5} d u

= \int \frac{2}{u ^{2} + 5} 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} + 5} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{5 ( 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}{5} \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}{5} arctan (v)

Substitute back

= 2 \cdot \frac{1}{5} arctan (\frac{x - 2}{5})

Simplify

2 \cdot \frac{1}{5} arctan (\frac{x - 2}{5}) : \frac{2}{5} arctan (\frac{1}{5} (x - 2))

= \frac{2}{5} arctan (\frac{1}{5} (x - 2))

Add a constant to the solution

= \frac{2}{5} arctan (\frac{1}{5} (x - 2)) + C

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