What is the integral from 1 to \sqrt[3]{5 of}(x^2+\sqrt[3]{x})/(x^2) ?

The integral from 1 to \sqrt[3]{5 of}(x^2+\sqrt[3]{x})/(x^2) is cube root of 5}+1/2-3/(2*5^{2/9)

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integral from 1 to \sqrt[3]{5 of}(x^2+\sqrt[3]{x})/(x^2)

\int_{1}^{35} \frac{x ^{2} + 3 x}{x ^{2}} d x

35 + \frac{1}{2} - \frac{3}{2 \cdot 5 ^{\frac{2}{9}}}

Decimal

1.16100 \dots

Solution steps

\int_{1}^{35} \frac{x ^{2} + 3 x}{x ^{2}} d x

Expand

\frac{x ^{2} + 3 x}{x ^{2}} : 1 + \frac{1}{x ^{\frac{5}{3}}}

= \int_{1}^{35} 1 + \frac{1}{x ^{\frac{5}{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}^{35} 1 d x + \int_{1}^{35} \frac{1}{x ^{\frac{5}{3}}} d x

\int_{1}^{35} 1 d x = 35 - 1

\int_{1}^{35} \frac{1}{x ^{\frac{5}{3}}} d x = - \frac{3}{2 \cdot 5 ^{\frac{2}{9}}} + \frac{3}{2}

= 35 - 1 - \frac{3}{2 \cdot 5 ^{\frac{2}{9}}} + \frac{3}{2}

Combine the fractions

- 1 + \frac{3}{2} : \frac{1}{2}

= 35 + \frac{1}{2} - \frac{3}{2 \cdot 5 ^{\frac{2}{9}}}

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What is the integral from 1 to \sqrt[3]{5 of}(x^2+\sqrt[3]{x})/(x^2) ?
The integral from 1 to \sqrt[3]{5 of}(x^2+\sqrt[3]{x})/(x^2) is cube root of 5}+1/2-3/(2*5^{2/9)