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

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

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

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

ln (\frac{1}{3} x^{2} + 6 x + x + 3) + C

Solution steps

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

Complete the square

x^{2} + 6 x : (x + 3)^{2} - 9

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

ln tan (arcsec (\frac{1}{3} (x + 3))) + sec (arcsec (\frac{1}{3} (x + 3))) : ln (\frac{1}{3} x^{2} + 6 x + x + 3)

= ln (\frac{1}{3} x^{2} + 6 x + x + 3)

Add a constant to the solution

= ln (\frac{1}{3} x^{2} + 6 x + x + 3) + C

\frac{d}{d x} (e^{1 x})

f (x) = 7 x - 2 x

\int \frac{2 x ^{2}}{2 x + 1} d x

\int \frac{2 x}{1 + 2 x} d x

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

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