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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{3}{t ( 1 + t ^{2} )} d t

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\int \frac{1}{g - k v ^{2}} d v

\int \frac{x ^{2}}{x ^{2} + y ^{2}} d x

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