What is the integral of 1/(sqrt(8x+x^2)) ?

The integral of 1/(sqrt(8x+x^2)) is ln(1/4 |sqrt(x^2+8x)+x+4|)+C

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integral of 1/(sqrt(8x+x^2))

\int \frac{1}{8 x + x ^{2}} d x

ln (\frac{1}{4} x^{2} + 8 x + x + 4) + C

Solution steps

\int \frac{1}{8 x + x ^{2}} d x

Complete the square

8 x + x^{2} : (x + 4)^{2} - 16

= \int \frac{1}{( x + 4 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 16} d u

= \int \frac{1}{u ^{2} - 16} 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}{4} (x + 4))) + sec (arcsec (\frac{1}{4} (x + 4)))

Simplify

ln tan (arcsec (\frac{1}{4} (x + 4))) + sec (arcsec (\frac{1}{4} (x + 4))) : ln (\frac{1}{4} x^{2} + 8 x + x + 4)

= ln (\frac{1}{4} x^{2} + 8 x + x + 4)

Add a constant to the solution

= ln (\frac{1}{4} x^{2} + 8 x + x + 4) + C

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What is the integral of 1/(sqrt(8x+x^2)) ?
The integral of 1/(sqrt(8x+x^2)) is ln(1/4 |sqrt(x^2+8x)+x+4|)+C