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Popular >

integral of (sqrt(x+6))/(x-3)

Solution

\int \frac{x + 6}{x - 3} d x

Solution

- 6 (\frac{1}{2} ln \frac{x + 6}{3} + 1 - \frac{1}{2} ln \frac{x + 6}{3} - 1 - \frac{1}{3} x + 6) + C

Solution steps

\int \frac{x + 6}{x - 3} d x

Apply u-substitution

: \int \frac{u}{u - 9} d u

= \int \frac{u}{u - 9} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} - 9} d v

= \int \frac{2 v ^{2}}{v ^{2} - 9} d v

Take the constant out:

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

= 2 \cdot \int \frac{v ^{2}}{v ^{2} - 9} d v

Factor

v^{2} - 9 : - (- v^{2} + 9)

= 2 \cdot \int \frac{v ^{2}}{- ( - v ^{2} + 9 )} d v

Take the constant out:

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

= 2 (- \int \frac{v ^{2}}{- v ^{2} + 9} d v)

Apply Integral Substitution

= 2 (- \int \frac{3 w ^{2}}{- w ^{2} + 1} d w)

Take the constant out:

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

= 2 (- 3 \cdot \int \frac{w ^{2}}{- w ^{2} + 1} d w)

\frac{w ^{2}}{- w ^{2} + 1} = \frac{1}{- w ^{2} + 1} - 1

= 2 (- 3 \cdot \int \frac{1}{- w ^{2} + 1} - 1 d w)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- 3 (\int \frac{1}{- w ^{2} + 1} d w - \int 1 d w))

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

\int 1 d w = w

= 2 (- 3 (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2} - w))

Substitute back

= 2 - 3 \frac{ln \frac{x + 6}{3} + 1}{2} - \frac{ln \frac{x + 6}{3} - 1}{2} - \frac{x + 6}{3}

Simplify

2 - 3 \frac{ln \frac{x + 6}{3} + 1}{2} - \frac{ln \frac{x + 6}{3} - 1}{2} - \frac{x + 6}{3} : - 6 (\frac{1}{2} ln \frac{x + 6}{3} + 1 - \frac{1}{2} ln \frac{x + 6}{3} - 1 - \frac{1}{3} x + 6)

= - 6 (\frac{1}{2} ln \frac{x + 6}{3} + 1 - \frac{1}{2} ln \frac{x + 6}{3} - 1 - \frac{1}{3} x + 6)

Add a constant to the solution

= - 6 (\frac{1}{2} ln \frac{x + 6}{3} + 1 - \frac{1}{2} ln \frac{x + 6}{3} - 1 - \frac{1}{3} x + 6) + C

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