Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
$$\int \frac{x^4 + 1}{x^4 + 1} dx =$$
\frac{1}{\sqrt{2}} \tan^{-1}\frac{x^2 - 1}{\sqrt{2}x} + C
\sin^{-1}\frac{x^2 + 1}{\sqrt{2}x} + C
\frac{1}{2}\log\frac{\sqrt{2}x + 1}{\sqrt{2}x - 1} + C
x^2 + \frac{1}{x^2} + C
Step-by-Step Solution
Key Concept: Divide numerator and denominator by $x^2$ and use the substitution $t = x - 1/x$ to convert to a standard arctangent form.
Rewrite $\int \frac{x^2+1}{x^4+1}dx = \int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$. Substitute $t = x - \frac{1}{x}$, giving $dt = (1+\frac{1}{x^2})dx$. Then $I = \int \frac{dt}{t^2+2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + c = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) + c$.
Correct Answer: 1