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

The integral of 1/(sqrt(16x^2+2)) is 1/4 ln|2sqrt(2)x+sqrt(1+8x^2)|+C

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

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

\frac{1}{4} ln 22 x + 1 + 8 x^{2} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4} d u

= \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{4}{2} x)

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

Simplify

\frac{1}{4} ln tan (arctan (\frac{4}{2} x)) + sec (arctan (\frac{4}{2} x)) : \frac{1}{4} ln 22 x + 1 + 8 x^{2}

= \frac{1}{4} ln 22 x + 1 + 8 x^{2}

Add a constant to the solution

= \frac{1}{4} ln 22 x + 1 + 8 x^{2} + C

\frac{d}{d x} (5 x + 9)

\frac{d}{d x} (cosh (x))

\int \frac{1}{y ^{2} y ^{2} + 25} d y

\frac{d}{d x} (7 cos^{2} (x))

f^{^{''}} (x) = 2, f^{^{'}} (2) = 5, f (2) = 10

What is the integral of 1/(sqrt(16x^2+2)) ?
The integral of 1/(sqrt(16x^2+2)) is 1/4 ln|2sqrt(2)x+sqrt(1+8x^2)|+C