What is the integral of 1/(xsqrt(x^4-36)) ?

The integral of 1/(xsqrt(x^4-36)) is 1/12 arctan(1/6 sqrt(x^4-36))+C

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integral of 1/(xsqrt(x^4-36))

\int \frac{1}{x x ^{4} - 36} d x

\frac{1}{12} arctan (\frac{1}{6} x^{4} - 36) + C

Solution steps

\int \frac{1}{x x ^{4} - 36} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 36 )} d u

= \int \frac{1}{2 ( u ^{2} + 36 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 36} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} \cdot \frac{1}{6} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{6} arctan (\frac{x ^{4} - 36}{6})

Simplify

\frac{1}{2} \cdot \frac{1}{6} arctan (\frac{x ^{4} - 36}{6}) : \frac{1}{12} arctan (\frac{1}{6} x^{4} - 36)

= \frac{1}{12} arctan (\frac{1}{6} x^{4} - 36)

Add a constant to the solution

= \frac{1}{12} arctan (\frac{1}{6} x^{4} - 36) + C

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\int x^{2} + 2 x + 1 d x

\frac{d y}{d x} = (\frac{x y + 5 x - y - 5}{x y - 2 x + 6 y - 12})

x y^{3} d x + e^{x^{2}} d y = 0

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

What is the integral of 1/(xsqrt(x^4-36)) ?
The integral of 1/(xsqrt(x^4-36)) is 1/12 arctan(1/6 sqrt(x^4-36))+C