Calculate following integration
$$\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx$$
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partial fractions? – gt6989b Mar 22 '13 at 12:40
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I don't think so – kalpeshmpopat Mar 22 '13 at 12:41
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This is not an elementary integral. – Mhenni Benghorbal Mar 22 '13 at 13:17
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@MhenniBenghorbal It actually is elementary, but the substitution I used is weird(I have seen this kind of integral before, which is the only reason I got it). Any easier substitution, you can think of? – Ishan Banerjee Mar 22 '13 at 13:35
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Sorry, I was evaluating the wrong integral. – Mhenni Benghorbal Mar 22 '13 at 13:56
2 Answers
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$\dfrac{1}{(x+1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}=\dfrac{1}{(x+1)^{2}(\dfrac{x+2}{x+1})^{\frac{5}{4}}}=\dfrac{1}{(x+1)^{2}(1+\dfrac{1}{x+1})^{\frac{5}{4}}}$ So, make a $\dfrac{1}{x+1}$ substitution and we get the answer.
Ishan Banerjee
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We have $$I=\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx=\int\frac{dx}{\left(\frac{x+1}{x+2}\right)^{3/4}(x+2)^2}$$ Now we pose $t=\frac{x+1}{x+2}$ then $x=\frac{1-2t}{t-1}$ and $dx=\frac{dt}{(t-1)^2}$, so $$I=\int t^{-3/4}dt=4t^{1/4}+C,$$ hence $$I=4\left(\frac{x+1}{x+2}\right)^{1/4}+C.$$