What is the integral of 1/(xsqrt(x+6)) ?

The integral of 1/(xsqrt(x+6)) is -1/(sqrt(6))(ln|sqrt((x+6)/6)+1|-ln|sqrt((x+6)/6)-1|)+C

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integral of 1/(xsqrt(x+6))

\int \frac{1}{x x + 6} d x

- \frac{1}{6} (ln \frac{x + 6}{6} + 1 - ln \frac{x + 6}{6} - 1) + C

Solution steps

\int \frac{1}{x x + 6} d x

Apply u-substitution

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

Take the constant out:

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

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

Apply Integral Substitution

= 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

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= 2 \cdot \frac{1}{6} - \frac{ln \frac{x + 6}{6} + 1}{2} - \frac{ln \frac{x + 6}{6} - 1}{2}

Simplify

2 \cdot \frac{1}{6} - \frac{ln \frac{x + 6}{6} + 1}{2} - \frac{ln \frac{x + 6}{6} - 1}{2} : - \frac{1}{6} (ln \frac{x + 6}{6} + 1 - ln \frac{x + 6}{6} - 1)

= - \frac{1}{6} (ln \frac{x + 6}{6} + 1 - ln \frac{x + 6}{6} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{x + 6}{6} + 1 - ln \frac{x + 6}{6} - 1) + C

\int \frac{( x ^{a} - x ^{b} ) ^{2}}{x} d x

d er i v a t i v e \frac{2}{t - 3}

s l o p e (- 2.7) (4.3)

3 (y^{4} + 1) d x + 4 x y^{3} d y = 0

\frac{d y}{d x} = \frac{2 x y}{1 + x ^{2}}

What is the integral of 1/(xsqrt(x+6)) ?
The integral of 1/(xsqrt(x+6)) is -1/(sqrt(6))(ln|sqrt((x+6)/6)+1|-ln|sqrt((x+6)/6)-1|)+C