What is the integral of (3+x)/(3-x) ?

The integral of (3+x)/(3-x) is 3-x-6ln|3-x|+C

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

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

3 - x - 6 ln ∣ 3 - x ∣ + C

Solution steps

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

Apply u-substitution

: \int - \frac{- u + 6}{u} d u

= \int - \frac{- u + 6}{u} d u

Take the constant out:

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

= - \int \frac{- u + 6}{u} d u

Expand

\frac{- u + 6}{u} : - 1 + \frac{6}{u}

= - \int - 1 + \frac{6}{u} d u

Apply the Sum Rule:

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

= - (- \int 1 d u + \int \frac{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= - (- u + 6 ln ∣ u ∣)

Substitute back

u = 3 - x

= - (- (3 - x) + 6 ln ∣ 3 - x ∣)

Simplify

= 3 - x - 6 ln ∣ 3 - x ∣

Add a constant to the solution

= 3 - x - 6 ln ∣ 3 - x ∣ + C

\frac{d}{d x} (\frac{3}{2} x^{\frac{1}{2}})

\int e^{5 x} sin (4 x) d x

\frac{d}{d x} ((5 x^{2} - 3) (x^{2} + x + 4))

\frac{\partial}{\partial y} (x + 5 y)

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 7 ) ^{\frac{3}{4}}})

What is the integral of (3+x)/(3-x) ?
The integral of (3+x)/(3-x) is 3-x-6ln|3-x|+C