What is the integral of 1/(sqrt(x^2+6x+5)) ?

The integral of 1/(sqrt(x^2+6x+5)) is ln(1/2 |sqrt(x^2+6x+5)+x+3|)+C

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integral of 1/(sqrt(x^2+6x+5))

\int \frac{1}{x ^{2} + 6 x + 5} d x

ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 5} d x

Complete the square

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

= \int \frac{1}{( x + 3 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 4} d u

= \int \frac{1}{u ^{2} - 4} 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}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3)))

Simplify

ln tan (arcsec (\frac{1}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3))) : ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3)

= ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3)

Add a constant to the solution

= ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + C

(8 (4 x)^{4} + 5)^{3}

\frac{d y}{d x} + y cos (x) = 6 cos (x), y (0) = 8

\int_{0}^{\frac{π}{22}} tan (11 x) d x

\frac{d y}{d x} + y^{3} x + 9 y = 0

x \to 7 lim (f (x) g (x))

What is the integral of 1/(sqrt(x^2+6x+5)) ?
The integral of 1/(sqrt(x^2+6x+5)) is ln(1/2 |sqrt(x^2+6x+5)+x+3|)+C