What is the integral of 1/(usqrt(6-u^2)) ?

The integral of 1/(usqrt(6-u^2)) is -1/(2sqrt(6))(ln|sqrt((6-u^2)/6)+1|-ln|sqrt((6-u^2)/6)-1|)+C

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integral of 1/(usqrt(6-u^2))

\int \frac{1}{u 6 - u ^{2}} d u

- \frac{1}{2 6} (ln \frac{6 - u ^{2}}{6} + 1 - ln \frac{6 - u ^{2}}{6} - 1) + C

Solution steps

\int \frac{1}{u 6 - u ^{2}} d u

Apply u-substitution

: \int - \frac{1}{- v ^{2} + 6} d v

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

Take the constant out:

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

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

Apply Integral Substitution

= - \int \frac{1}{6 ( - w ^{2} + 1 )} d w

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= - \frac{1}{6} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= - \frac{1}{6} \frac{ln \frac{6 - u ^{2}}{6} + 1}{2} - \frac{ln \frac{6 - u ^{2}}{6} - 1}{2}

Simplify

- \frac{1}{6} \frac{ln \frac{6 - u ^{2}}{6} + 1}{2} - \frac{ln \frac{6 - u ^{2}}{6} - 1}{2} : - \frac{1}{2 6} (ln \frac{6 - u ^{2}}{6} + 1 - ln \frac{6 - u ^{2}}{6} - 1)

= - \frac{1}{2 6} (ln \frac{6 - u ^{2}}{6} + 1 - ln \frac{6 - u ^{2}}{6} - 1)

Add a constant to the solution

= - \frac{1}{2 6} (ln \frac{6 - u ^{2}}{6} + 1 - ln \frac{6 - u ^{2}}{6} - 1) + C

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}})

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f (x) = x^{3} + 1, (1, 2)

\int \frac{5 - e ^{- x}}{e ^{- x}} d x

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

What is the integral of 1/(usqrt(6-u^2)) ?
The integral of 1/(usqrt(6-u^2)) is -1/(2sqrt(6))(ln|sqrt((6-u^2)/6)+1|-ln|sqrt((6-u^2)/6)-1|)+C