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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\large\int\frac{ x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}dx$ $(Hint:Put\;x=z^4)$

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  • If $f(x)$ is substituted by t; then $f'(x)dx=f'(t)dt$ and $ \int f(x)dx=\int f(t)dt$
  • $\large\int \frac{dx}{(x+a)}=\log |x+a|+c$
Step 1:
Let $I=\large\int\frac{ x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}dx$
Put $x=z^4$
Differentiating with respect to z we get
$dx=4z^3 .dz$
On substituting for x and dx in I we get,
Therefore $I=\large\int \frac{(z^4)^{1/2}}{1+(z^4)^{3/4}}.4z^3dz$
$I=\large\int \frac{z^2}{1+z^3}.4 z^3.dz$
$=4 \Large\int \frac{z^5}{1+z^3}.dz$$=4 \int \Large\frac{z^3.z^2.dz}{1+z^3}$
Put $z^3=t$
differentiating with respect to t
$3z^2dz=dt => z^2dz=dt/3$
Substituting for z and dz we get,
$I=4 \int \large\frac{t.(dt/3)}{1+t}=\frac{4}{3} \int \frac{t\; dt}{1+t}$
Add and subtract 1 to the numerator
Step 2:
$I=\large\frac{4}{3} \int \frac{t+1-1}{1+t}dt$
$I=\large\frac{4}{3} \bigg[\int dt -\int \frac{dt}{t+1}\bigg]$
On integrating we get,
$I=\large\frac{4}{3} \bigg[t- \log |t+1|\bigg]+c$
Substituting for t we get,
$\large\frac{4}{3}\bigg[z^3-\log |z^3+1|\bigg]$+c
$x=z^4 =>x^{1/4}=z$
Therefore $ z^3=x^{3/4}$
Now substituting for z we get
$\large\frac{4}{3} \bigg[x^{3/4}-\log |x^{3/4}+1|\bigg]+c$
answered Apr 15, 2013 by meena.p
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