What is the integral of (4x^2+x+2)/(x+1) ?

The integral of (4x^2+x+2)/(x+1) is 2x^2-3x+5ln|x+1|-5+C

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integral of (4x^2+x+2)/(x+1)

\int \frac{4 x ^{2} + x + 2}{x + 1} d x

2 x^{2} - 3 x + 5 ln ∣ x + 1 ∣ - 5 + C

Solution steps

\int \frac{4 x ^{2} + x + 2}{x + 1} d x

Expand

\frac{4 x ^{2} + x + 2}{x + 1} : \frac{4 x ^{2}}{x + 1} + \frac{x}{x + 1} + \frac{2}{x + 1}

Apply the Sum Rule:

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

= \int \frac{4 x ^{2}}{x + 1} d x + \int \frac{x}{x + 1} d x + \int \frac{2}{x + 1} d x

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

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

Simplify

4 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + x + 1 - ln ∣ x + 1 ∣ + 2 ln ∣ x + 1 ∣ : 2 x^{2} - 3 x + 5 ln ∣ x + 1 ∣ - 5

= 2 x^{2} - 3 x + 5 ln ∣ x + 1 ∣ - 5

Add a constant to the solution

= 2 x^{2} - 3 x + 5 ln ∣ x + 1 ∣ - 5 + C

What is the integral of (4x^2+x+2)/(x+1) ?
The integral of (4x^2+x+2)/(x+1) is 2x^2-3x+5ln|x+1|-5+C