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If $$f(x) =\frac {x+2}{2x+3}$$ then the integral $$ \int{\sqrt{\frac{f(x)}{x^2}}}\,dx$$ is equal to?

The expression is way beyond my comprehension, no idea in this case.

mathemather
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Akshay Pratap Singh
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1 Answers1

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Let $$\displaystyle I = \int\frac{1}{x}\sqrt{\frac{x+2}{2x+3}}dx\;,$$ Now Let $\displaystyle \frac{x+2}{2x+3}=t^2\Rightarrow \frac{(2x+3)+1}{2x+3}=2t^2\;,$

Then $$\displaystyle -\frac{1}{(2x+3)^2}dx = 2tdt$$ and $$\displaystyle 2x+3=\frac{1}{2t^2-1}\Rightarrow 2x=\frac{4-6t^2}{2t^2-1}$$

So Integral $$\displaystyle I = \int\frac{2}{2x}\cdot \sqrt{\frac{x+2}{2x+3}}dx = -\int\frac{2t\cdot 2t\cdot (2t^2-1)}{(4-6t^2)\cdot (2t^2-1)^2}dt = 2\int\frac{t^2}{(3t^2-2)(2t^2-1)}dt$$

So $$\displaystyle I = 2\int \left[\frac{2(2t^2-1)-(3t^2-2)}{(3t^2-2)\cdot (2^t2-1)}\right]dt = 4\int\frac{1}{3t^2-2}dt-2\int\frac{1}{2t^2-1}dt$$

So we get $$\displaystyle I = \frac{4}{3}\int\frac{1}{t^2-\left(\sqrt{\frac{2}{3}}\right)^2}dt-\int\frac{1}{t^2-\left(\frac{1}{\sqrt{2}}\right)^2}dt$$

Now Using the formula $$\displaystyle \int\frac{1}{x^2-a^2}dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+\mathcal{C}$$

Perhaps someone have better method then that.

juantheron
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  • how can be $\sqrt{x^2}=x$? – Dr. Sonnhard Graubner Aug 11 '15 at 18:40
  • actually $\sqrt{x^2} = |x|;,$ Which is $+x,x>0$ and $-x,x\leq 0$ – juantheron Aug 11 '15 at 18:42
  • above I have calculate only for $x>0$ – juantheron Aug 11 '15 at 18:42
  • FWIW, this is the form that Wolfram Alpha gives, so juantheron's answer is likely the furthest simplification. $$\frac{\sqrt{\frac{2 + x}{3 + 2 x}} \sqrt{3 + 2 x} }{3 \sqrt{2} \sqrt{2 + x}}\left[{2 \sqrt{3} \ln{x} + 3 \ln\left(7 + 4 x + 2 \sqrt{2} \sqrt{2 + x} \sqrt{3 + 2 x}\right) - 2 \sqrt{3} \ln\left(12 + 7 x + 2 \sqrt{6} \sqrt{2 + x} \sqrt{3 + 2 x}\right)}\right]+C$$ – Jam Aug 11 '15 at 19:10