What is the integral of 1/(sqrt(x^2+12x)) ?

The integral of 1/(sqrt(x^2+12x)) is ln(1/6 |sqrt(x^2+12x)+x+6|)+C

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integral of 1/(sqrt(x^2+12x))

\int \frac{1}{x ^{2} + 12 x} d x

ln (\frac{1}{6} x^{2} + 12 x + x + 6) + C

Solution steps

\int \frac{1}{x ^{2} + 12 x} d x

Complete the square

x^{2} + 12 x : (x + 6)^{2} - 36

= \int \frac{1}{( x + 6 ) ^{2} - 36} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 36} d u

= \int \frac{1}{u ^{2} - 36} 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}{6} (x + 6))) + sec (arcsec (\frac{1}{6} (x + 6)))

Simplify

ln tan (arcsec (\frac{1}{6} (x + 6))) + sec (arcsec (\frac{1}{6} (x + 6))) : ln (\frac{1}{6} x^{2} + 12 x + x + 6)

= ln (\frac{1}{6} x^{2} + 12 x + x + 6)

Add a constant to the solution

= ln (\frac{1}{6} x^{2} + 12 x + x + 6) + C

y^{^{'}} = 25 + y^{2}

\frac{s + 2}{( s ^{2} + 4 )}

\frac{d}{d t} (\frac{1}{3} t^{3})

y^{^{''}} + 25 y = t e^{6 t}

\frac{d y}{d x}, x y = 3

What is the integral of 1/(sqrt(x^2+12x)) ?
The integral of 1/(sqrt(x^2+12x)) is ln(1/6 |sqrt(x^2+12x)+x+6|)+C