What is the integral from 3 to 5 of (2x^2+x+4)/(x-1) ?

The integral from 3 to 5 of (2x^2+x+4)/(x-1) is 7ln(2)+22

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integral from 3 to 5 of (2x^2+x+4)/(x-1)

\int_{3}^{5} \frac{2 x ^{2} + x + 4}{x - 1} d x

7 ln (2) + 22

Decimal Notation

26.85203 \dots

Solution steps

\int_{3}^{5} \frac{2 x ^{2} + x + 4}{x - 1} d x

Expand

\frac{2 x ^{2} + x + 4}{x - 1} : \frac{2 x ^{2}}{x - 1} + \frac{x}{x - 1} + \frac{4}{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_{3}^{5} \frac{2 x ^{2}}{x - 1} d x + \int_{3}^{5} \frac{x}{x - 1} d x + \int_{3}^{5} \frac{4}{x - 1} d x

\int_{3}^{5} \frac{2 x ^{2}}{x - 1} d x = 2 (ln (2) + 10)

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

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

= 2 (ln (2) + 10) + 2 + ln (2) + 4 ln (2)

Simplify

2 (ln (2) + 10) + 2 + ln (2) + 4 ln (2) : 7 ln (2) + 22

= 7 ln (2) + 22

\int_{1}^{e} \frac{( 1 + ln ( x ) ) ^{2}}{x} d x

\int \frac{tan ( ln ( x ) )}{x} d x

(3 x^{3} + x^{- 3}) (x + 3) (x^{2} - 5)

f^{^{'}} (x) = 2 x^{3} - x

y = c e^{\frac{- x ^{3}}{3}} + 1

What is the integral from 3 to 5 of (2x^2+x+4)/(x-1) ?
The integral from 3 to 5 of (2x^2+x+4)/(x-1) is 7ln(2)+22