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Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$

I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$

But this does not seem to work here.I am stuck.What should i do?

Vinod Kumar Punia
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3 Answers3

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First step is put $x = u+\dfrac{1}{2}\Rightarrow \sqrt{x^2+(x-1)^2}=\sqrt{\left(u+\dfrac{1}{2}\right)^2+\left(u-\dfrac{1}{2}\right)^2}=\sqrt{2u^2+\dfrac{1}{2}}=\sqrt{2}\cdot \sqrt{u^2+\dfrac{1}{4}}$. Second step is to put $u =\dfrac{1}{2}\tan \theta$. I think we can go further at this point.

DeepSea
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First write $$x - 1 = x(2x - 1) - (2x^2 - 2x + 1).$$ Then our integrand becomes: $$\frac{x-1}{x^2 \sqrt{2x^2 - 2x + 1}} = \frac{2x-1}{x\sqrt{2x^2 - 2x + 1}} - \frac{\sqrt{2x^2 - 2x + 1}}{x^2}.$$ Now with the choice $$u = \sqrt{2x^2 - 2x + 1}, \quad v = \frac{1}{x},$$ we have $$u' = \frac{2x-1}{\sqrt{2x^2-2x+1}}, \quad v' = -\frac{1}{x^2},$$ hence the integrand is of the form $$u'v + uv' = (uv)'.$$ It immediately follows that an antiderivative is $$\int \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \, dx = uv + C = \frac{\sqrt{2x^2 - 2x + 1}}{x} + C.$$ The rest is straightforward.

heropup
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  • Unique approach +1,how does splitting $x-1$ in that way click to you.@heropup – Vinod Kumar Punia Jan 08 '16 at 07:27
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    The idea is to try to see if the integrand is amenable to being written in the form of a product rule. In order to do this, we try to write it as the sum/difference of two terms. This suggests trying to "pull out" of the numerator the quadratic polynomial that occurs in the square root, to see if its derivative is remaining; i.e., note that $(2x^2 - 2x + 1)' = 2(2x-1)$, which suggests such a trick may work. – heropup Jan 08 '16 at 07:33
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you write the integral as $$I=\int \frac{1-\frac{1}{x}}{x^2\sqrt{1+(1-\frac{1}{x})^2}}$$ now put $1-\frac{1}{x}=y$

Ekaveera Gouribhatla
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