What is the integral of 1/(sqrt(7+8x^2)) ?

The integral of 1/(sqrt(7+8x^2)) is 1/(2sqrt(2))ln|(2sqrt(2))/(sqrt(7))x+sqrt(1/7 (7+8x^2))|+C

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integral of 1/(sqrt(7+8x^2))

\int \frac{1}{7 + 8 x ^{2}} d x

\frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2}) + C

Solution steps

\int \frac{1}{7 + 8 x ^{2}} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= \frac{1}{2 2} \cdot \int sec (u) d u

Use the common integral:

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

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

Substitute back

u = arctan (\frac{2 2}{7} x)

= \frac{1}{2 2} ln tan (arctan (\frac{2 2}{7} x)) + sec (arctan (\frac{2 2}{7} x))

Simplify

\frac{1}{2 2} ln tan (arctan (\frac{2 2}{7} x)) + sec (arctan (\frac{2 2}{7} x)) : \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2})

= \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2})

Add a constant to the solution

= \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2}) + C

n = 0 \sum \infty n! (x - 4)^{n}

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

\frac{x ^{2}}{x - 1}

f (x) = \frac{1}{ln ( x )}

What is the integral of 1/(sqrt(7+8x^2)) ?
The integral of 1/(sqrt(7+8x^2)) is 1/(2sqrt(2))ln|(2sqrt(2))/(sqrt(7))x+sqrt(1/7 (7+8x^2))|+C