What is the integral of 4/(xsqrt(16x^2-25)) ?

The integral of 4/(xsqrt(16x^2-25)) is 4/5 arcsec(4/5 x)+C

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integral of 4/(xsqrt(16x^2-25))

\int \frac{4}{x 16 x ^{2} - 25} d x

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

Solution steps

\int \frac{4}{x 16 x ^{2} - 25} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x 16 x ^{2} - 25} d x

Apply Trigonometric Substitution

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

= 4 \cdot \int \frac{1}{5} d u

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{5} x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 4/(xsqrt(16x^2-25)) ?
The integral of 4/(xsqrt(16x^2-25)) is 4/5 arcsec(4/5 x)+C