What is the integral of (7x^3+6x)/(x-1) ?

The integral of (7x^3+6x)/(x-1) is 27x+13ln|x-1|+7/3 (x-1)^3+21/2 (x-1)^2-27+C

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integral of (7x^3+6x)/(x-1)

\int \frac{7 x ^{3} + 6 x}{x - 1} d x

27 x + 13 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 27 + C

Solution steps

\int \frac{7 x ^{3} + 6 x}{x - 1} d x

Expand

\frac{7 x ^{3} + 6 x}{x - 1} : \frac{7 x ^{3}}{x - 1} + \frac{6 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{7 x ^{3}}{x - 1} d x + \int \frac{6 x}{x - 1} d x

\int \frac{7 x ^{3}}{x - 1} d x = 7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣)

\int \frac{6 x}{x - 1} d x = 6 (x - 1 + ln ∣ x - 1 ∣)

= 7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 6 (x - 1 + ln ∣ x - 1 ∣)

Simplify

7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 6 (x - 1 + ln ∣ x - 1 ∣) : 27 x + 13 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 27

= 27 x + 13 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 27

Add a constant to the solution

= 27 x + 13 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 27 + C

d er i v a t i v e 3 x^{- 2 x}

\frac{\partial}{\partial x} (\frac{m c T}{m T})

t \to 0 lim (\frac{1}{t} - \frac{1}{e ^{t} - 1})

\int x 3 - x d x

\int_{- 24}^{0} \frac{1}{1 - x} d x

What is the integral of (7x^3+6x)/(x-1) ?
The integral of (7x^3+6x)/(x-1) is 27x+13ln|x-1|+7/3 (x-1)^3+21/2 (x-1)^2-27+C