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

The integral from 4 to 5 of (4x^2+x-2)/(x-3) is 37ln(2)+31

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

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

37 ln (2) + 31

Decimal

56.64644 \dots

Solution steps

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

Expand

\frac{4 x ^{2} + x - 2}{x - 3} : \frac{4 x ^{2}}{x - 3} + \frac{x}{x - 3} - \frac{2}{x - 3}

Apply the Sum Rule:

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

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

\int_{4}^{5} \frac{4 x ^{2}}{x - 3} d x = 4 (9 ln (2) + \frac{15}{2})

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

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

= 4 (9 ln (2) + \frac{15}{2}) + 1 + 3 ln (2) - 2 ln (2)

Simplify

4 (9 ln (2) + \frac{15}{2}) + 1 + 3 ln (2) - 2 ln (2) : 37 ln (2) + 31

= 37 ln (2) + 31

\int_{0}^{1} \frac{x}{x ^{2} + 3 x + 2} d x

\int_{0}^{1} x^{3} - x d x

\int_{0}^{2} \frac{1}{1 + e ^{1 - x}} d x

\int_{0}^{\frac{π}{2}} x cos^{2} (x) d x

\int_{0}^{3} x^{2} - 1 d x

What is the integral from 4 to 5 of (4x^2+x-2)/(x-3) ?
The integral from 4 to 5 of (4x^2+x-2)/(x-3) is 37ln(2)+31