What is the integral of 1/(sqrt(4x^2-7)) ?

The integral of 1/(sqrt(4x^2-7)) is 1/2 ln|sqrt(1/7 (4x^2-7))+2/(sqrt(7))x|+C

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integral of 1/(sqrt(4x^2-7))

\int \frac{1}{4 x ^{2} - 7} d x

\frac{1}{2} ln \frac{1}{7} (4 x^{2} - 7) + \frac{2}{7} x + C

Solution steps

\int \frac{1}{4 x ^{2} - 7} 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 = arcsec (\frac{2}{7} x)

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

Simplify

\frac{1}{2} ln tan (arcsec (\frac{2}{7} x)) + sec (arcsec (\frac{2}{7} x)) : \frac{1}{2} ln \frac{1}{7} (4 x^{2} - 7) + \frac{2}{7} x

= \frac{1}{2} ln \frac{1}{7} (4 x^{2} - 7) + \frac{2}{7} x

Add a constant to the solution

= \frac{1}{2} ln \frac{1}{7} (4 x^{2} - 7) + \frac{2}{7} x + C

\int_{0}^{t} 9 e^{s} sin (t - s) d s

f (x) = (2 - x) e^{x}

x \to 0 + lim (2)

t

x \to \infty lim (- (x^{2}))

What is the integral of 1/(sqrt(4x^2-7)) ?
The integral of 1/(sqrt(4x^2-7)) is 1/2 ln|sqrt(1/7 (4x^2-7))+2/(sqrt(7))x|+C