What is the integral of 1/(sqrt(x^2+10x)) ?

The integral of 1/(sqrt(x^2+10x)) is ln(1/5 |sqrt(x^2+10x)+x+5|)+C

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integral of 1/(sqrt(x^2+10x))

\int \frac{1}{x ^{2} + 10 x} d x

ln (\frac{1}{5} x^{2} + 10 x + x + 5) + C

Solution steps

\int \frac{1}{x ^{2} + 10 x} d x

Complete the square

x^{2} + 10 x : (x + 5)^{2} - 25

= \int \frac{1}{( x + 5 ) ^{2} - 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 25} d u

= \int \frac{1}{u ^{2} - 25} 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}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5)))

Simplify

ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))) : ln (\frac{1}{5} x^{2} + 10 x + x + 5)

= ln (\frac{1}{5} x^{2} + 10 x + x + 5)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} + 10 x + x + 5) + C

\frac{d y}{d x} = - \frac{( e ^{x} )}{8}

\int - 2 e^{2 t} d t

\int \frac{1}{1 + 17 x} d x

\int (x sin (x)) d x

y = \frac{2 x}{4 + x ^{2}}

What is the integral of 1/(sqrt(x^2+10x)) ?
The integral of 1/(sqrt(x^2+10x)) is ln(1/5 |sqrt(x^2+10x)+x+5|)+C