Mathematics | VITEEE 2016 | Discussion Page

$\int\frac{1}{x^{2}(x^{4}+1)^{3/4}}dx$ is equal to

$(1+\frac{1}{x^{4}})^{1/4}+C$

$(x^{4}+1)^{1/4}+C$

$(1-\frac{1}{x^{4}})^{1/4}+C$

$-(1+\frac{1}{x^{4}})^{1/4}+C$

Option D

$\int \frac{dx}{x^{2}(x^{4}-1)^{3/4}}=\int \frac{dx}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{3/4}}$

Put $1+\frac{1}{x^{4}}=t\Rightarrow-\frac{4}{x^{5}}dx=dt$

So, integral is

$I= -\frac{1}{4}\int\frac{dt}{t^{3/4}}=-t^{\frac{1}{4}}+c=-\left(1+\frac{1}{x^{4}}\right)^{1/4}+c$

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