What is the integral from 0 to 2 of (4x+3)/(x^4) ?

The integral from 0 to 2 of (4x+3)/(x^4) is diverges

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integral from 0 to 2 of (4x+3)/(x^4)

\int_{0}^{2} \frac{4 x + 3}{x ^{4}} d x

diverges

Solution steps

\int_{0}^{2} \frac{4 x + 3}{x ^{4}} d x

Expand

\frac{4 x + 3}{x ^{4}} : \frac{4}{x ^{3}} + \frac{3}{x ^{4}}

= \int_{0}^{2} \frac{4}{x ^{3}} + \frac{3}{x ^{4}} 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_{0}^{2} \frac{4}{x ^{3}} d x + \int_{0}^{2} \frac{3}{x ^{4}} d x

\int_{0}^{2} \frac{4}{x ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int (5 x^{3} - 7 x^{2} + 2 x + 1) d x

\frac{d}{d x} (30 + \frac{1}{x + 1} - 3 x^{2})

\frac{d}{d x} (\frac{1}{5 x ^{2} + 2 x})

\frac{5 s + 3}{( s - 1 ) ( s ^{2} + 2 s + 5 )}

\frac{\partial}{\partial y} (x^{2} + e^{x y})

What is the integral from 0 to 2 of (4x+3)/(x^4) ?
The integral from 0 to 2 of (4x+3)/(x^4) is diverges