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

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

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

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

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

Solution steps

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

Complete the square

4 x^{2} + 4 x + 2 : 4 (x + \frac{1}{2})^{2} + 1

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

Apply u-substitution

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

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

Simplify

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

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

Add a constant to the solution

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

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