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

The integral of (5x^3+4x)/(x-1) is 19x+9ln|x-1|+5/3 (x-1)^3+15/2 (x-1)^2-19+C

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

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

19 x + 9 ln ∣ x - 1 ∣ + \frac{5}{3} (x - 1)^{3} + \frac{15}{2} (x - 1)^{2} - 19 + C

Solution steps

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

Expand

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

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

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

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

Simplify

5 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 4 (x - 1 + ln ∣ x - 1 ∣) : 19 x + 9 ln ∣ x - 1 ∣ + \frac{5}{3} (x - 1)^{3} + \frac{15}{2} (x - 1)^{2} - 19

= 19 x + 9 ln ∣ x - 1 ∣ + \frac{5}{3} (x - 1)^{3} + \frac{15}{2} (x - 1)^{2} - 19

Add a constant to the solution

= 19 x + 9 ln ∣ x - 1 ∣ + \frac{5}{3} (x - 1)^{3} + \frac{15}{2} (x - 1)^{2} - 19 + C

\int \frac{x - 4}{x} d x

y^{^{'}} = \frac{7 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

(- 8, - 3), (- 3, 4)

f (x) = \frac{5}{x}, at x = 1

\frac{1}{2 s + 3}

What is the integral of (5x^3+4x)/(x-1) ?
The integral of (5x^3+4x)/(x-1) is 19x+9ln|x-1|+5/3 (x-1)^3+15/2 (x-1)^2-19+C