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

The integral from 1 to 5 of 4/x-4/(x^2) is 4ln(5)-16/5

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

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

4 ln (5) - \frac{16}{5}

Solution steps

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

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

\int_{1}^{5} \frac{4}{x ^{2}} d x = \frac{16}{5}

= 4 ln (5) - \frac{16}{5}

\int_{4}^{8} x^{2} d x

\int_{a}^{x} \frac{f ( t )}{t ^{2}} d t

\int_{0}^{3} arctan (x) d x

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

\int_{1}^{5} 2 x - 1 d x

What is the integral from 1 to 5 of 4/x-4/(x^2) ?
The integral from 1 to 5 of 4/x-4/(x^2) is 4ln(5)-16/5