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# Integrate the function$\frac{1}{x^\frac{1}{2}+x^\frac{1}{3}}\qquad[Hint:\frac{1}{x^\frac{1}{2}+x^\frac{1}{3}}=\frac{1}{x^\frac{1}{3}\bigg(1+x^\frac{1}{6}\bigg)},put\;x=t^6]$

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• (i)If f(x)=t, then f'(x)=dt.then if $I=\int f(x)dx,$it can be written as $\int t.dt$
• (ii)$\int\frac{1}{x+a}dx=log |x+a|+c$
Given $I=\int\large\frac{1}{x^{1/2}+x^{1/3}}dx$

This can be written as $\int\large\frac{1}{x^{1/3}(1+x^{1/6})}dx$

Let $x=t^6$ on differentiating we get $dx=6t^5dx$

On substituting for t and dt we get,

$I=\int \large\frac{6t^5dt}{(t^6)^{1/3}(1+(t)}$

On simplifying we get,

$I=6\int\large\frac{t^3dt}{(1+t)}$

Since it is an improper rational function, let us divide,

Hence the function is $(t^2-t+1)-(\frac{1}{1+t})$

Therefore $I=6\int (t^2-t+1)dt-6\int\frac{1}{1+t}dt$

On integrating we get,

$\frac{6t^3}{3}-\frac{6t^2}{2}+6t+6log|1+t|+c$

Substituting for t we get,

$2(x^{1/6})^3-3(x^{1/6})^2+6(x^{1/6})-6log |1+x^{1/6}|+c$

$I=2\sqrt x-3x^{1/3}+6x^{1/6}-6log(1+x^{1/6})+c$

answered Feb 15, 2013 by
edited Aug 8, 2013