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

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

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

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

ln (3) + \frac{4}{3}

Decimal

2.43194 \dots

Solution steps

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

Expand

\frac{x ^{2} + x - 1}{x ^{2}} : 1 + \frac{1}{x} - \frac{1}{x ^{2}}

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

\int_{1}^{3} 1 d x = 2

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

\int_{1}^{3} \frac{1}{x ^{2}} d x = \frac{2}{3}

= 2 + ln (3) - \frac{2}{3}

Combine the fractions

2 - \frac{2}{3} : \frac{4}{3}

= ln (3) + \frac{4}{3}

\int_{0}^{1} x \cdot arctan (x) d x

\int_{- 2}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x

\int_{4}^{8} \frac{x}{x ^{2} - 15} d x

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

\int_{3}^{3} x d x

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