Question

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

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

Answer 1

The correct option is D $-(1+\frac{1}{x^4}{)}^{1/4}+c$

$\int \frac{1}{x^{25}}(1{+}^1{/}_{x}4{)}^{{}^3{/}_4}dx$

$\frac{1}{x^4}=t$

$\frac{-4}{x^5}dx=dt$

$\int \frac{1}{4}\frac{dt}{(1+t{)}^{{}^3{/}_4}}=-(1+t{)}^{\frac{1}{4}}+c$

$=-(1+t{)}^{{}^1{/}_4}+c$

$=-(1{+}^1{/}_{x^4}{)}^{{}^1{/}_4}+c$

$\int \frac{1}{x^2(x^4+1{)}^{{}^3{/}_4}}dx=(1+\frac{1}{x^4}{)}^{\frac{1}{4}}+c$