Integral Calculus Question 119
Question: $ \int{\frac{\sqrt{x}}{1+\sqrt[4]{x^{3}}}}dx $ is equal to
Options:
A) $ \frac{4}{3}[ 1+{x^{3/4}}+\log (1+{x^{3/4}}) ]+C $
B) $ \frac{4}{3}[ 1+{x^{3/4}}-\log (1+{x^{3/4}}) ]+C $
C) $ \frac{4}{3}[ 1-{x^{3/4}}+\log (1+{x^{3/4}}) ]+C $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Put $ x=z^{4}\Rightarrow dx=4z^{3}dz $
$ \therefore \int{\frac{\sqrt[2]{x}}{1+\sqrt[4]{x^{3}}}dx=\int{\frac{z^{2}.4z^{3}}{1+z^{3}}dz=4\int{\frac{z^{3}.z^{2}}{z^{3}+1}dz}}} $ $ =\frac{4}{3}\int{\frac{(y-1)}{y}dy} $ $ [ Puttingz^{3}+1=y\Rightarrow z^{2}dz=\frac{1}{3}dy ] $ $ =\frac{4}{3}(y-logy)+C $ $ =\frac{4}{3}[ 1+{x^{3/4}}-\log (1+{x^{3/4}}) ]+C $
BETA
