What is the integral of 2/(x^3sqrt(x^4-81)) ?

The integral of 2/(x^3sqrt(x^4-81)) is (sqrt(x^4-81))/(81x^2)+C

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integral of 2/(x^3sqrt(x^4-81))

\int \frac{2}{x ^{3} x ^{4} - 81} d x

\frac{x ^{4} - 81}{81 x ^{2}} + C

Solution steps

\int \frac{2}{x ^{3} x ^{4} - 81} 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 ^{3} x ^{4} - 81} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} u ^{2} - 81} d u

= 2 \cdot \int \frac{1}{2 u ^{2} u ^{2} - 81} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} u ^{2} - 81} d u

Apply Trigonometric Substitution

: \int \frac{1}{81 sec ( v )} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{81 sec ( v )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{81} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= 2 \cdot \frac{1}{2} \cdot \frac{1}{81} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 2 \cdot \frac{1}{2} \cdot \frac{1}{81} sin (v)

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{81} sin (arcsec (\frac{1}{9} x^{2}))

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{81} sin (arcsec (\frac{1}{9} x^{2})) : \frac{x ^{4} - 81}{81 x ^{2}}

= \frac{x ^{4} - 81}{81 x ^{2}}

Add a constant to the solution

= \frac{x ^{4} - 81}{81 x ^{2}} + C

\frac{d}{d t} (e^{- a t} cos (wt))

\int \frac{x ^{2} 2 ^{x} + x}{x ^{2}} d x

\frac{d}{d x} (\frac{( 3 x ^{2} + 1 ) ^{3}}{( 2 x ^{2} - 1 ) ^{4}})

f^{^{'}} (x) = a^{2 x}

n = 1 \sum \infty (\frac{n - 2}{n})^{n}

What is the integral of 2/(x^3sqrt(x^4-81)) ?
The integral of 2/(x^3sqrt(x^4-81)) is (sqrt(x^4-81))/(81x^2)+C