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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{2}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 10

\frac{d}{d x} (\frac{6 x}{1 + 3 x ^{2}})

x \to 0 lim (\frac{1}{x} ln (1 + x))

\int_{0}^{π} t^{2} e^{- s t} d t

y = e^{x}, y = e^{- x}, x = 1

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