What is the integral of 1/(sqrt(x^2+20x)) ?

The integral of 1/(sqrt(x^2+20x)) is ln(1/10 |sqrt(x^2+20x)+x+10|)+C

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integral of 1/(sqrt(x^2+20x))

\int \frac{1}{x ^{2} + 20 x} d x

ln (\frac{1}{10} x^{2} + 20 x + x + 10) + C

Solution steps

\int \frac{1}{x ^{2} + 20 x} d x

Complete the square

x^{2} + 20 x : (x + 10)^{2} - 100

= \int \frac{1}{( x + 10 ) ^{2} - 100} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 100} d u

= \int \frac{1}{u ^{2} - 100} 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}{10} (x + 10))) + sec (arcsec (\frac{1}{10} (x + 10)))

Simplify

ln tan (arcsec (\frac{1}{10} (x + 10))) + sec (arcsec (\frac{1}{10} (x + 10))) : ln (\frac{1}{10} x^{2} + 20 x + x + 10)

= ln (\frac{1}{10} x^{2} + 20 x + x + 10)

Add a constant to the solution

= ln (\frac{1}{10} x^{2} + 20 x + x + 10) + C

\frac{\partial}{\partial x} (2 x - 3 y^{2} + z^{2})

\frac{d}{d x} (\frac{14}{x ^{3}})

\frac{\partial}{\partial x} (x^{\frac{x}{y}} cos (\frac{x}{y}))

n = 1 \sum \infty \frac{1}{( 5 n - 4 ) ^{4}}

\int - 0.4 x^{2} + 1000 d x

What is the integral of 1/(sqrt(x^2+20x)) ?
The integral of 1/(sqrt(x^2+20x)) is ln(1/10 |sqrt(x^2+20x)+x+10|)+C