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

The integral of (3x^4-2x+5)/(x+1) is (3x^4)/4-x^3+(3x^2)/2-5x+10ln|x+1|-2+C

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integral of (3x^4-2x+5)/(x+1)

\int \frac{3 x ^{4} - 2 x + 5}{x + 1} d x

\frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 5 x + 10 ln ∣ x + 1 ∣ - 2 + C

Solution steps

\int \frac{3 x ^{4} - 2 x + 5}{x + 1} d x

Expand

\frac{3 x ^{4} - 2 x + 5}{x + 1} : \frac{3 x ^{4}}{x + 1} - \frac{2 x}{x + 1} + \frac{5}{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{3 x ^{4}}{x + 1} d x - \int \frac{2 x}{x + 1} d x + \int \frac{5}{x + 1} d x

\int \frac{3 x ^{4}}{x + 1} d x = \frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 3 x + 3 ln ∣ x + 1 ∣

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

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

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

Simplify

\frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 3 x + 3 ln ∣ x + 1 ∣ - 2 (x + 1 - ln ∣ x + 1 ∣) + 5 ln ∣ x + 1 ∣ : \frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 5 x + 10 ln ∣ x + 1 ∣ - 2

= \frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 5 x + 10 ln ∣ x + 1 ∣ - 2

Add a constant to the solution

= \frac{3 x ^{4}}{4} - x^{3} + \frac{3 x ^{2}}{2} - 5 x + 10 ln ∣ x + 1 ∣ - 2 + C

\frac{d}{d x} (A x)

\frac{d}{d y} (1 - x y cos (π y))

\frac{x ^{2}}{40}

x y^{^{'}} = e^{x} - y

f (t) = \frac{1}{20} t^{5} e^{2 t}

What is the integral of (3x^4-2x+5)/(x+1) ?
The integral of (3x^4-2x+5)/(x+1) is (3x^4)/4-x^3+(3x^2)/2-5x+10ln|x+1|-2+C