What is the integral of 1/(sqrt(4x^2+12x+5)) ?

The integral of 1/(sqrt(4x^2+12x+5)) is 1/2 ln|(sqrt(4x^2+12x+5))/2+x+3/2 |+C

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integral of 1/(sqrt(4x^2+12x+5))

\int \frac{1}{4 x ^{2} + 12 x + 5} d x

\frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2} + C

Solution steps

\int \frac{1}{4 x ^{2} + 12 x + 5} d x

Complete the square

4 x^{2} + 12 x + 5 : 4 (x + \frac{3}{2})^{2} - 4

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arcsec (x + \frac{3}{2})) + sec (arcsec (x + \frac{3}{2}))

Simplify

\frac{1}{2} ln tan (arcsec (x + \frac{3}{2})) + sec (arcsec (x + \frac{3}{2})) : \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2}

= \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2}

Add a constant to the solution

= \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2} + C

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What is the integral of 1/(sqrt(4x^2+12x+5)) ?
The integral of 1/(sqrt(4x^2+12x+5)) is 1/2 ln|(sqrt(4x^2+12x+5))/2+x+3/2 |+C