What is the integral of (7x^3+4x)/(x-1) ?

The integral of (7x^3+4x)/(x-1) is 25x+11ln|x-1|+7/3 (x-1)^3+21/2 (x-1)^2-25+C

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integral of (7x^3+4x)/(x-1)

\int \frac{7 x ^{3} + 4 x}{x - 1} d x

25 x + 11 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 25 + C

Solution steps

\int \frac{7 x ^{3} + 4 x}{x - 1} d x

Expand

\frac{7 x ^{3} + 4 x}{x - 1} : \frac{7 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{7 x ^{3}}{x - 1} d x + \int \frac{4 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{4 x}{x - 1} d x = 4 (x - 1 + ln ∣ x - 1 ∣)

= 7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 4 (x - 1 + ln ∣ x - 1 ∣)

Simplify

7 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) + 4 (x - 1 + ln ∣ x - 1 ∣) : 25 x + 11 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 25

= 25 x + 11 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 25

Add a constant to the solution

= 25 x + 11 ln ∣ x - 1 ∣ + \frac{7}{3} (x - 1)^{3} + \frac{21}{2} (x - 1)^{2} - 25 + C

y^{^{''}} + 25 y = - 40 sin (5 t)

x^{2} y^{^{'}} - 2 x y = y^{4}

\frac{d}{d x} (cos^{2} (tan (x)))

x \to 0 lim (\frac{5 x}{x ^{3}})

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 6})

What is the integral of (7x^3+4x)/(x-1) ?
The integral of (7x^3+4x)/(x-1) is 25x+11ln|x-1|+7/3 (x-1)^3+21/2 (x-1)^2-25+C