Question

$\int \frac{1-x^7}{x(1+x^7)}dx$ equals -

Answer 1

$\ln|x| - \frac{2}{7}\ln|1+x^7| + C$

Answer 2

To evaluate the integral $\int \frac{1-x^7}{x(1+x^7)}dx$, we can use the method of substitution followed by partial fraction decomposition.

1. Substitution: Let $t = x^7$.
   Differentiating both sides with respect to $x$:
   $dt = 7x^6 dx$
   This implies $dx = \frac{dt}{7x^6}$.

   Now, substitute $t$ and $dx$ into the integral: $\int \frac{1-x^7}{x(1+x^7)}dx = \int \frac{1-t}{x(1+t)} \left(\frac{dt}{7x^6}\right)$ $= \int \frac{1-t}{7x^7(1+t)} dt$ Since $x^7 = t$, we can replace $x^7$ in the denominator: $= \int \frac{1-t}{7t(1+t)} dt$ $= \frac{1}{7} \int \frac{1-t}{t(1+t)} dt$

2. Partial Fraction Decomposition: Now, we need to decompose the rational function $\frac{1-t}{t(1+t)}$ into partial fractions.
   Let: $\frac{1-t}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}$ Multiply both sides by $t(1+t)$:
   $1-t = A(1+t) + Bt$ To find $A$: Set $t=0$:
   $1-0 = A(1+0) + B(0)$
   $1 = A$

   To find $B$: Set $t=-1$:
   $1-(-1) = A(1-1) + B(-1)$
   $2 = 0 - B$
   $B = -2$

   So, the partial fraction decomposition is: $\frac{1-t}{t(1+t)} = \frac{1}{t} - \frac{2}{1+t}$

3. Integration: Substitute the partial fractions back into the integral: $\frac{1}{7} \int \left( \frac{1}{t} - \frac{2}{1+t} \right) dt$ Integrate each term: $= \frac{1}{7} \left( \int \frac{1}{t} dt - \int \frac{2}{1+t} dt \right)$ $= \frac{1}{7} \left( \ln|t| - 2\ln|1+t| \right) + C$

4. Substitute back $t=x^7$: $= \frac{1}{7} \left( \ln|x^7| - 2\ln|1+x^7| \right) + C$ Using the logarithm property $\ln(a^b) = b\ln(a)$: $= \frac{1}{7} \left( 7\ln|x| - 2\ln|1+x^7| \right) + C$ Distribute the $\frac{1}{7}$: $= \ln|x| - \frac{2}{7}\ln|1+x^7| + C$