# Evaluate:$\int\limits_0^{16} \frac{x^{1/4}}{1+x^{1/2}}dx$

## 1 Answer

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)If we substitute a function f(x) as t, then $f'(x)dx=dt$ hence the function $\int f(x)dx=\int t.dt$
• (iii)$\int \frac {dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1} (x/a)+c$
Given $I=\int\limits_0^{16} \frac{x^{1/4}}{1+x^{1/2}}dx$

Let $x^{1/4}=t.$ on differentiating w.r.t x

$\frac{1}{4} x^{-3/4}dx=dt$

$=\frac{1}{4} t^{-3} dx =dt \qquad =\frac{1}{4t^3}dx=dt$

As we substitute for t the limit changes

Therefore $dx=4t^3dt$

when $x=0,t=0, x=16,t=2$

Substituting for t and dt,

$I=\int \limits_0^2 \large\frac{t(4t^3)dt}{1+t^2}$

$=4 \int _0^2 \large\frac{t^4 dt}{1+t^2}$

This is an improper rational function,

hence on dividing we get,

Therefore $\frac{t^4}{1+t^2}=(t^2-1)+\frac{1}{1+t^2}$

Therefore $4 \int \limits_0^2\frac{t^4}{1+t^2}dt=4 \int \limits_0^2[(t^2-1)+\frac{1}{1+t^2}]dt$

On seperating the terms,

$I=\int \limits_0^2 t^2-\int \limits_0^2 dt+\int \limits_0^2 \frac{dt}{1+t^2}$

$\large\frac{dt}{1+t^2}$ is of the form $\frac {dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1} (x/a)$

On integrating we get

Therefore $I=4 \bigg\{[t^3/3]_0^2-[t]_0^2+[\tan ^{-1}(t)]_0^2 \bigg\}$

On applying the limits,

$=4 \bigg\{(\frac{8}{3}-0)-(2-0)+\tan ^{-1}(2)\bigg\}$

$=\frac {32}{3}-8+4 \tan ^{-1}(2)$

$=\frac{8}{3}+4 \tan ^{-1}(2)$

answered Mar 15, 2013 by

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