What is the integral of 1/(sqrt(x^2+8x+3)) ?

The integral of 1/(sqrt(x^2+8x+3)) is ln|sqrt(1/13 (x^2+8x+3))+1/(sqrt(13))(x+4)|+C

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integral of 1/(sqrt(x^2+8x+3))

\int \frac{1}{x ^{2} + 8 x + 3} d x

ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4) + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 3} d x

Complete the square

x^{2} + 8 x + 3 : (x + 4)^{2} - 13

= \int \frac{1}{( x + 4 ) ^{2} - 13} d x

Apply u-substitution

= \int \frac{1}{u ^{2} - 13} d u

Apply Trigonometric Substitution

= \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}{13} (x + 4))) + sec (arcsec (\frac{1}{13} (x + 4)))

Simplify

ln tan (arcsec (\frac{1}{13} (x + 4))) + sec (arcsec (\frac{1}{13} (x + 4))) : ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4)

= ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4)

Add a constant to the solution

= ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4) + C

What is the integral of 1/(sqrt(x^2+8x+3)) ?
The integral of 1/(sqrt(x^2+8x+3)) is ln|sqrt(1/13 (x^2+8x+3))+1/(sqrt(13))(x+4)|+C