Question

The value of $\int \limits \frac {dx}{x^2(x^4+1)^{3/4}}$ is

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

Answer 1

The Correct Option is D

To solve the integral $\int \frac {dx}{x^2(x^4+1)^{3/4}}$, we start by applying the substitution method:

1. Let $x^4 + 1 = t$. Then, the derivative is $4x^3 \, dx = dt$.
2. This substitution allows us to express $dx$ as $\frac{dt}{4x^3}$.
3. Substitute $x^4$ from $t = x^4 + 1 \Rightarrow x^4 = t - 1$.
4. The integral becomes: $$ \int \frac{dx}{x^2(t)^{3/4}} = \int \frac{\frac{dt}{4x^3}}{x^2(t)^{3/4}} $$
5. Since $x^4 = t - 1$, we have $x^3 = (t - 1)^{3/4}$.
6. Substituting $x^3 = \frac{t - 1}{x}$ simplifies further to: $$ \int \frac{dt}{4(t-1)^{3/4} \cdot (t)^{3/4}} $$
7. Simplify the integral: $$ \int \frac{dt}{4 (t(t-1))^{3/4}} $$

This integral becomes straightforward with another substitution or can be recognized as leading to a power form:

1. Make a direct integral calculation as this type involves known integrations: $$ F(t) = -\frac{(t)^{1/4}}{(t-1)^{1/4}} + C $$
2. Substitute back $t = x^4 + 1$, giving: $$ -\left(\frac{x^4+1}{x^4}\right)^{1/4} + C $$

Thus, the correct answer is:

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

Hence, the option $- \bigg (\frac {x^4+1}{x^4} \bigg )^ \frac {1}{4}+c$ is correct.