Question

The value of \(\int\frac{1+x^4}{1+x^6}dx\) is

• A. tan^-1 x + tan^-1 x³ + C
• B. tan^-1 x + \(\frac{1}{3}\)tan^-1 x³ + C
• C. tan^-1 x - \(\frac{1}{3}\)tan^-1 x³ + C
• D. tan^-1 x + \(\frac{1}{3}\)tan^-1 x² + C

Answer 1

The Correct Option is B

We need to find the value of \(\int \frac{1+x^4}{1+x^6} dx\).

We can rewrite the integrand as:

\(\int \frac{1+x^4}{1+x^6} dx = \int \frac{1+x^4}{(1+x^2)(1-x^2+x^4)} dx\)

\(\int \frac{1+x^4}{(1+x^2)(1-x^2+x^4)} dx= \int \frac{1+x^4}{1+x^6} dx\)

Now, we try to write the numerator in terms of \(1-x^2+x^4\) and \(1+x^2\) to simplify the integral:

We have \(1+x^4= (1-x^2+x^4) +x^2\). Hence \(\int \frac{1+x^4}{1+x^6} dx = \int \frac{(1-x^2+x^4)+ x^2}{(1+x^2)(1-x^2+x^4)} dx\).

Splitting the integral into two terms, \(\int \frac{1+x^4}{1+x^6} dx = \int \frac{(1-x^2+x^4)}{(1+x^2)(1-x^2+x^4)} dx + \int \frac{ x^2}{(1+x^2)(1-x^2+x^4)} dx\).

\(\int \frac{1+x^4}{1+x^6} dx = \int \frac{1}{1+x^2} dx + \int \frac{x^2}{1+x^6} dx\).

\(\int \frac{1+x^4}{1+x^6} dx = \arctan(x)+ \int \frac{x^2}{1+(x^3)^2} dx\).

Let \(u=x^3\) so \(du=3x^2dx\).

\(\int \frac{1+x^4}{1+x^6} dx = \arctan(x)+ \frac{1}{3}\int \frac{du}{1+u^2}\).

\(\int \frac{1+x^4}{1+x^6} dx = \arctan(x)+ \frac{1}{3}\arctan(u) + C\).

\(\int \frac{1+x^4}{1+x^6} dx = \arctan(x)+ \frac{1}{3}\arctan(x^3) + C\).

Therefore, the correct option is (B) \(\tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C\).

Answer 2

We want to find $ \int \frac{1+x^4}{1+x^6} dx $.

Notice that $ 1+x^6 = (1+x^2)(1-x^2+x^4) $. Let's rewrite the fraction as:

$$ \frac{1+x^4}{1+x^6} = \frac{1+x^4}{(1+x^2)(1-x^2+x^4)} = \frac{1+x^4}{1+x^2} \cdot \frac{1}{1-x^2+x^4}. $$

Observe that $ 1+x^4 = (1-x^2+x^4) + x^2 $. 

Thus:

$$ \frac{1+x^4}{1+x^6} = \frac{1-x^2+x^4+x^2}{(1+x^2)(1-x^2+x^4)} = \frac{1-x^2+x^4}{(1+x^2)(1-x^2+x^4)} + \frac{x^2}{(1+x^2)(1-x^2+x^4)}. $$

Simplify further:

$$ \frac{1+x^4}{1+x^6} = \frac{1}{1+x^2} + \frac{x^2}{(1+x^2)(1-x^2+x^4)}. $$

Now, let us consider an alternative approach. Rewrite the integrand:

$$ \frac{1+x^4}{1+x^6} = \frac{1+x^4}{(1+x^2)(1-x^2+x^4)} = \frac{1}{1+x^2} - \frac{x^2}{1+x^6}. $$

Integrating term by term:

$$ \int \frac{1+x^4}{1+x^6} dx = \int \frac{1}{1+x^2} dx - \int \frac{x^2}{1+x^6} dx. $$

The first term integrates as:

$$ \int \frac{1}{1+x^2} dx = \tan^{-1} x. $$

For the second term, substitute $ u = x^3 $, so $ du = 3x^2 dx $ and $ x^2 dx = \frac{1}{3} du $. Thus:

$$ \int \frac{x^2}{1+x^6} dx = \frac{1}{3} \int \frac{1}{1+u^2} du = \frac{1}{3} \tan^{-1} u = \frac{1}{3} \tan^{-1} x^3. $$

Combining these results:

$$ \int \frac{1+x^4}{1+x^6} dx = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C. $$