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

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

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integral of 1/((x+3)sqrt(x^2+6x))

\int \frac{1}{( x + 3 ) x ^{2} + 6 x} d x

\frac{1}{3} arcsec (\frac{1}{3} x + 1) + C

Solution steps

\int \frac{1}{( x + 3 ) x ^{2} + 6 x} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 9} d u

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x + 3))

Simplify

\frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x + 3)) : \frac{1}{3} arcsec (\frac{1}{3} x + 1)

= \frac{1}{3} arcsec (\frac{1}{3} x + 1)

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} x + 1) + C

f (x) = \frac{x}{7} - 1

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

\int 14 x - x^{2} d x

\frac{d}{d x} (x^{\frac{- 2}{5}})

f (x) = 2 x cos (x^{2})

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