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

The integral of x/(xsqrt(81x^2-81)) is 1/9 ln|sqrt(x^2-1)+x|+C

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

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

\frac{1}{9} ln x^{2} - 1 + x + C

Solution steps

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

Simplify

\frac{x}{x 81 x ^{2} - 81} : \frac{1}{9 x ^{2} - 1}

= \int \frac{1}{9 x ^{2} - 1} d x

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{x ^{2} - 1} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \frac{1}{9} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{9} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (x)

= \frac{1}{9} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣

Simplify

\frac{1}{9} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ : \frac{1}{9} ln x^{2} - 1 + x

= \frac{1}{9} ln x^{2} - 1 + x

Add a constant to the solution

= \frac{1}{9} ln x^{2} - 1 + x + C

\frac{d}{d x} ((ln (2 x + 1))^{\frac{1}{2}})

\int tan^{2} (2 x) sec^{2} (2 x) d x

\frac{d}{d x} (3 x - 5 x)

\int x (x + 7)^{4} d x

(1 + x^{2}) \frac{d y}{d x} - x y = 0

What is the integral of x/(xsqrt(81x^2-81)) ?
The integral of x/(xsqrt(81x^2-81)) is 1/9 ln|sqrt(x^2-1)+x|+C