What is the integral of ((x+3)(x-1))/(4(x+1)) ?

The integral of ((x+3)(x-1))/(4(x+1)) is (x^2)/8+1/4 x-ln|x+1|+C

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integral of ((x+3)(x-1))/(4(x+1))

\int \frac{( x + 3 ) ( x - 1 )}{4 ( x + 1 )} d x

\frac{x ^{2}}{8} + \frac{1}{4} x - ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{( x + 3 ) ( x - 1 )}{4 ( x + 1 )} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{( x + 3 ) ( x - 1 )}{x + 1} d x

Long division

\frac{( x + 3 ) ( x - 1 )}{x + 1} : x + 1 - \frac{4}{x + 1}

= \frac{1}{4} \cdot \int x + 1 - \frac{4}{x + 1} d x

Apply the Sum Rule:

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

= \frac{1}{4} (\int x d x + \int 1 d x - \int \frac{4}{x + 1} d x)

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

= \frac{1}{4} (\frac{x ^{2}}{2} + x - 4 ln ∣ x + 1 ∣)

Simplify

\frac{1}{4} (\frac{x ^{2}}{2} + x - 4 ln ∣ x + 1 ∣) : \frac{x ^{2}}{8} + \frac{1}{4} x - ln ∣ x + 1 ∣

= \frac{x ^{2}}{8} + \frac{1}{4} x - ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{x ^{2}}{8} + \frac{1}{4} x - ln ∣ x + 1 ∣ + C

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What is the integral of ((x+3)(x-1))/(4(x+1)) ?
The integral of ((x+3)(x-1))/(4(x+1)) is (x^2)/8+1/4 x-ln|x+1|+C