What is the integral from 1 to 2 of (6x+5)/(x^2+x) ?

The integral from 1 to 2 of (6x+5)/(x^2+x) is 4ln(2)+ln(3)

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integral from 1 to 2 of (6x+5)/(x^2+x)

\int_{1}^{2} \frac{6 x + 5}{x ^{2} + x} d x

4 ln (2) + ln (3)

Decimal

3.87120 \dots

Solution steps

\int_{1}^{2} \frac{6 x + 5}{x ^{2} + x} d x

Take the partial fraction of

\frac{6 x + 5}{x ^{2} + x} : \frac{5}{x} + \frac{1}{x + 1}

= \int_{1}^{2} \frac{5}{x} + \frac{1}{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_{1}^{2} \frac{5}{x} d x + \int_{1}^{2} \frac{1}{x + 1} d x

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

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

= 5 ln (2) + ln (3) - ln (2)

Simplify

= 4 ln (2) + ln (3)

\int_{- 1}^{3} (x + 3) d x

\int_{0}^{1} 4 (2 x + 1)^{3} d x

\int_{- 3}^{1} (6 x^{2} - 5 x + 2) d x

\int_{0}^{b} (2^{x} + x^{2}) d x

\int_{0}^{5} 5^{5 x} d x

What is the integral from 1 to 2 of (6x+5)/(x^2+x) ?
The integral from 1 to 2 of (6x+5)/(x^2+x) is 4ln(2)+ln(3)