What is the integral from 1 to 6 of (4x^2-3)/(x^3) ?

The integral from 1 to 6 of (4x^2-3)/(x^3) is 4ln(6)-35/24

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integral from 1 to 6 of (4x^2-3)/(x^3)

\int_{1}^{6} \frac{4 x ^{2} - 3}{x ^{3}} d x

4 ln (6) - \frac{35}{24}

Decimal

5.70870 \dots

Solution steps

\int_{1}^{6} \frac{4 x ^{2} - 3}{x ^{3}} d x

Expand

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

= \int_{1}^{6} \frac{4}{x} - \frac{3}{x ^{3}} 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}^{6} \frac{4}{x} d x - \int_{1}^{6} \frac{3}{x ^{3}} d x

\int_{1}^{6} \frac{4}{x} d x = 4 ln (6)

\int_{1}^{6} \frac{3}{x ^{3}} d x = \frac{35}{24}

= 4 ln (6) - \frac{35}{24}

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What is the integral from 1 to 6 of (4x^2-3)/(x^3) ?
The integral from 1 to 6 of (4x^2-3)/(x^3) is 4ln(6)-35/24