What is the integral of 1/(xsqrt(64x^2-5)) ?

The integral of 1/(xsqrt(64x^2-5)) is 1/(sqrt(5))arcsec(8/(sqrt(5))x)+C

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integral of 1/(xsqrt(64x^2-5))

\int \frac{1}{x 64 x ^{2} - 5} d x

\frac{1}{5} arcsec (\frac{8}{5} x) + C

Solution steps

\int \frac{1}{x 64 x ^{2} - 5} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \int \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{8}{5} x)

= \frac{1}{5} \cdot 1 \cdot arcsec (\frac{8}{5} x)

Simplify

= \frac{1}{5} arcsec (\frac{8}{5} x)

Add a constant to the solution

= \frac{1}{5} arcsec (\frac{8}{5} x) + C

\int_{- 1}^{1} (0 - y^{2} + 1) d y

12 y^{^{''}} - 12 y^{^{'}} + 30 y = 0

\int \frac{3 x ^{3}}{4 + x ^{8}} d x

\frac{d x}{d y} = 3 x^{2} y^{3}

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} d x

What is the integral of 1/(xsqrt(64x^2-5)) ?
The integral of 1/(xsqrt(64x^2-5)) is 1/(sqrt(5))arcsec(8/(sqrt(5))x)+C