Solution to Finding the Integral of (x-4)/x^2 from 1 to 2

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Homework Statement


Find the integral of (x-4)/x^2 dx from 1 to 2.


Homework Equations


Do I divide the denominator? I got 1/x-4/x^2 from 1 to 2. But what to do next?


The Attempt at a Solution


The answer is ln(2)-2.
 
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Nerd10 said:

Homework Statement


Find the integral of (x-4)/x^2 dx from 1 to 2.

Homework Equations


Do I divide the denominator? I got 1/x-4/x^2 from 1 to 2. But what to do next?

The Attempt at a Solution


The answer is ln(2)-2.

There are multiple methods to solve this integral, but the method you chose (dividing through by the denominator) is probably the easiest. The integral of two functions added together is the sum of the individual integrals of the functions: ##\int (f(x)+g(x))dx=\int f(x)dx+\int g(x)dx##
 
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Nerd10 said:
I got 1/x-4/x^2 from 1 to 2. But what to do next
Are you able to integrate 1/x^2 ? If so, then multiply the result by -4 and you'll have this problem half-solved.
 
Can it be 1/x^2=x^-2? And I integrate from there?
 
Never mind. I got it. Thanks for the help.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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