What is the integral of (5x^3+7x)/(x-1) ?

The integral of (5x^3+7x)/(x-1) is (5x^3)/3+(5x^2)/2+12x+12ln|x-1|+C

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integral of (5x^3+7x)/(x-1)

\int \frac{5 x ^{3} + 7 x}{x - 1} d x

\frac{5 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 12 x + 12 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{5 x ^{3} + 7 x}{x - 1} d x

Long division

\frac{5 x ^{3} + 7 x}{x - 1} : 5 x^{2} + 5 x + 12 + \frac{12}{x - 1}

= \int 5 x^{2} + 5 x + 12 + \frac{12}{x - 1} d x

Apply the Sum Rule:

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

= \int 5 x^{2} d x + \int 5 x d x + \int 12 d x + \int \frac{12}{x - 1} d x

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 12 d x = 12 x

\int \frac{12}{x - 1} d x = 12 ln ∣ x - 1 ∣

= \frac{5 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 12 x + 12 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{5 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 12 x + 12 ln ∣ x - 1 ∣ + C

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

y = 2 x arccos (x) - 2 1 - x^{2}

n = 0 \sum \infty \frac{( n )}{n + 1}

\frac{y}{x ^{2}} \frac{d y}{d x} + e^{2 x^{3} + y^{2}} = 0

x \to \infty lim (\frac{4}{x} - 12)

What is the integral of (5x^3+7x)/(x-1) ?
The integral of (5x^3+7x)/(x-1) is (5x^3)/3+(5x^2)/2+12x+12ln|x-1|+C