What is the integral of 2/(xsqrt(25x^2-64)) ?

The integral of 2/(xsqrt(25x^2-64)) is 1/4 arcsec(5/8 x)+C

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integral of 2/(xsqrt(25x^2-64))

\int \frac{2}{x 25 x ^{2} - 64} d x

\frac{1}{4} arcsec (\frac{5}{8} x) + C

Solution steps

\int \frac{2}{x 25 x ^{2} - 64} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x 25 x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int \frac{1}{8} d u

= 2 \cdot \int \frac{1}{8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{8} \cdot 1 \cdot u

Substitute back

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

= 2 \cdot \frac{1}{8} \cdot 1 \cdot arcsec (\frac{5}{8} x)

Simplify

2 \cdot \frac{1}{8} \cdot 1 \cdot arcsec (\frac{5}{8} x) : \frac{1}{4} arcsec (\frac{5}{8} x)

= \frac{1}{4} arcsec (\frac{5}{8} x)

Add a constant to the solution

= \frac{1}{4} arcsec (\frac{5}{8} x) + C

\int \frac{1}{( x ^{2} + 8 x + 25 ) ^{\frac{3}{2}}} d x

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

\int \frac{sin ^{2} ( π ) x}{cos ^{2} ( π ) x} d x

\int z arcsec (z) d z

What is the integral of 2/(xsqrt(25x^2-64)) ?
The integral of 2/(xsqrt(25x^2-64)) is 1/4 arcsec(5/8 x)+C