Question

The intergral $\int \frac{dx}{x^2(x^4+1{)}^{3/4}}$ eauals:

• A. $-(x^4+1{)}^{1/4}+c$
• B. $-(\frac{x^4+1}{x^4}{)}^{1/4}+c$
• C. $(\frac{x^4+1}{x4}{)}^{1/4}+c$
• D.
  $(x^4+1{)}^{1/4}+c$

Answer 1

The correct option is B

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

I$=\int \frac{dx}{x^2(x^4+1{)}^{3/4}}=\int \frac{dx}{x^5(1+x^{-4}{)}^{3/4}}Let{x}^{-4}=y\Rightarrow -4x^{-5}dx=dy\Rightarrow dx=\frac{-1}{4}{x}^{5}dy=I=\frac{-1}{4}\int \frac{x^{5}dy}{x^5(1+y{)}^{\frac{3}{4}}}=\frac{-1}{4}\int \frac{dy}{(1+y{)}^{3/4}}=\frac{-1}{4}\times 4(1+y{)}^{1/4}=-(1+x^{-4}{)}^{1/4}+c=-(\frac{x^4+1}{x^4}{)}^{1/4}+c$