What is the integral of 1/(xsqrt(x^4-16)) ?

The integral of 1/(xsqrt(x^4-16)) is 1/8 arctan(1/4 sqrt(x^4-16))+C

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integral of 1/(xsqrt(x^4-16))

\int \frac{1}{x x ^{4} - 16} d x

\frac{1}{8} arctan (\frac{1}{4} x^{4} - 16) + C

Solution steps

\int \frac{1}{x x ^{4} - 16} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 16 )} d u

= \int \frac{1}{2 ( u ^{2} + 16 )} 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} + 16} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{4} - 16}{4})

Simplify

\frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{4} - 16}{4}) : \frac{1}{8} arctan (\frac{1}{4} x^{4} - 16)

= \frac{1}{8} arctan (\frac{1}{4} x^{4} - 16)

Add a constant to the solution

= \frac{1}{8} arctan (\frac{1}{4} x^{4} - 16) + C

\frac{d}{d x} (ln (x) + ln (sin (y)))

\frac{d}{d x} (- 7 x)

(x + 1) \frac{d y}{d x} + y = ln (x), y (1) = 15

\frac{x ^{2} - 1}{x ^{2} + 4}

What is the integral of 1/(xsqrt(x^4-16)) ?
The integral of 1/(xsqrt(x^4-16)) is 1/8 arctan(1/4 sqrt(x^4-16))+C