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

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

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

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

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

Solution steps

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

Complete the square

2 x^{2} + 4 x + 5 : 2 (x + 1)^{2} + 3

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

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 3} d u

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2} + 6}{sin ^{4} ( x )})

f (x) = - 9 e^{x^{- 9}}

\frac{d}{d x} (- \frac{2}{1 - 2 x})

x \to - 2 lim (3 x^{2} + 4 x - 2)

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