What is the integral of 5/(xsqrt(16x^2-81)) ?

The integral of 5/(xsqrt(16x^2-81)) is 5/9 arcsec(4/9 x)+C

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integral of 5/(xsqrt(16x^2-81))

\int \frac{5}{x 16 x ^{2} - 81} d x

\frac{5}{9} arcsec (\frac{4}{9} x) + C

Solution steps

\int \frac{5}{x 16 x ^{2} - 81} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x 16 x ^{2} - 81} d x

Apply Trigonometric Substitution

: \int \frac{1}{9} d u

= 5 \cdot \int \frac{1}{9} d u

Take the constant out:

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

= 5 \cdot \frac{1}{9} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{9} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{9} x)

= 5 \cdot \frac{1}{9} \cdot 1 \cdot arcsec (\frac{4}{9} x)

Simplify

5 \cdot \frac{1}{9} \cdot 1 \cdot arcsec (\frac{4}{9} x) : \frac{5}{9} arcsec (\frac{4}{9} x)

= \frac{5}{9} arcsec (\frac{4}{9} x)

Add a constant to the solution

= \frac{5}{9} arcsec (\frac{4}{9} x) + C

8 ln (\frac{5 x + 7}{3 x ^{2} - 6 x})

f (x) = sin (x^{2} + 1)

f (x) = \frac{x ^{2}}{1 + x}

y y^{^{'}} = x^{2}

\int a^{x} d x

What is the integral of 5/(xsqrt(16x^2-81)) ?
The integral of 5/(xsqrt(16x^2-81)) is 5/9 arcsec(4/9 x)+C