What is the integral of 1/(sqrt(x^2+4x+3)) ?

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

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integral of 1/(sqrt(x^2+4x+3))

\int \frac{1}{x ^{2} + 4 x + 3} d x

ln x^{2} + 4 x + 3 + x + 2 + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 3} d x

Complete the square

x^{2} + 4 x + 3 : (x + 2)^{2} - 1

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

Apply u-substitution

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

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

Simplify

ln ∣ tan (arcsec (x + 2)) + sec (arcsec (x + 2)) ∣ : ln x^{2} + 4 x + 3 + x + 2

= ln x^{2} + 4 x + 3 + x + 2

Add a constant to the solution

= ln x^{2} + 4 x + 3 + x + 2 + C

\int_{2}^{4} - x^{2} + 6 x - 8 d x

x \to - 8 - lim (\frac{x + 7}{x + 8})

cos (\frac{x}{2})

x \to 4 lim (\frac{x ^{2} - 8}{x ^{4}})

\int x ln (ln (x)) d x

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