What is the integral of 1/(sqrt(4x^2+12x+16)) ?

The integral of 1/(sqrt(4x^2+12x+16)) is 1/2 ln| 2/(sqrt(7))x+3/(sqrt(7))+sqrt((4x^2+12x+16)/7)|+C

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integral of 1/(sqrt(4x^2+12x+16))

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

\frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7} + C

Solution steps

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

Complete the square

4 x^{2} + 12 x + 16 : 4 (x + \frac{3}{2})^{2} + 7

= \int \frac{1}{4 ( x + \frac{3}{2} ) ^{2} + 7} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 7} d u

= \int \frac{1}{4 u ^{2} + 7} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

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

Take the constant out:

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

= \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 (arctan (\frac{2}{7} (x + \frac{3}{2}))) + sec (arctan (\frac{2}{7} (x + \frac{3}{2})))

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{7} (x + \frac{3}{2}))) + sec (arctan (\frac{2}{7} (x + \frac{3}{2}))) : \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7}

= \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7}

Add a constant to the solution

= \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7} + C

(7, - 5), (- 7, - 5)

\frac{d}{d x} (y (e^{x} - 1))

\int (\frac{x}{5})^{n} d x

\int_{- 1}^{1} 2^{x} d x

\frac{d}{d x} (2 + \frac{3}{1 + ( x + 1 ) ^{2}})

What is the integral of 1/(sqrt(4x^2+12x+16)) ?
The integral of 1/(sqrt(4x^2+12x+16)) is 1/2 ln| 2/(sqrt(7))x+3/(sqrt(7))+sqrt((4x^2+12x+16)/7)|+C