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

The integral from 1 to 2 of (x^3+3x^6)/(x^4) is ln(2)+7

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

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

ln (2) + 7

Decimal

7.69314 \dots

Solution steps

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

Expand

\frac{x ^{3} + 3 x ^{6}}{x ^{4}} : \frac{1}{x} + 3 x^{2}

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

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

\int_{1}^{2} 3 x^{2} d x = 7

= ln (2) + 7

What is the integral from 1 to 2 of (x^3+3x^6)/(x^4) ?
The integral from 1 to 2 of (x^3+3x^6)/(x^4) is ln(2)+7