What is the integral of 1/(sqrt(x^2+18x)) ?

The integral of 1/(sqrt(x^2+18x)) is ln(1/9 |sqrt(x^2+18x)+x+9|)+C

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integral of 1/(sqrt(x^2+18x))

\int \frac{1}{x ^{2} + 18 x} d x

ln (\frac{1}{9} x^{2} + 18 x + x + 9) + C

Solution steps

\int \frac{1}{x ^{2} + 18 x} d x

Complete the square

x^{2} + 18 x : (x + 9)^{2} - 81

= \int \frac{1}{( x + 9 ) ^{2} - 81} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 81} d u

= \int \frac{1}{u ^{2} - 81} 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}{9} (x + 9))) + sec (arcsec (\frac{1}{9} (x + 9)))

Simplify

ln tan (arcsec (\frac{1}{9} (x + 9))) + sec (arcsec (\frac{1}{9} (x + 9))) : ln (\frac{1}{9} x^{2} + 18 x + x + 9)

= ln (\frac{1}{9} x^{2} + 18 x + x + 9)

Add a constant to the solution

= ln (\frac{1}{9} x^{2} + 18 x + x + 9) + C

\frac{d}{d t} (tan (t))

x \to 3 lim (x - 2)

\frac{\partial}{\partial x} (L - x \cdot cos (b))

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{- x}, y (0) = 0, y^{^{'}} (1) = 2

\int y cos (y) d y

What is the integral of 1/(sqrt(x^2+18x)) ?
The integral of 1/(sqrt(x^2+18x)) is ln(1/9 |sqrt(x^2+18x)+x+9|)+C