Question

If $\int \frac{(2x^2+1)}{(x^2-4)(x^2-1)} = log\left[\left(\frac{x+1}{x-1}\right)^a \left(\frac{x-2}{x+2}\right)^b \right]+c$

Answer 1

a=1/2, b=3/4

Answer 2

The problem requires us to evaluate the given integral and then compare it with the provided logarithmic form to find the values of 'a' and 'b'.

1. Partial Fraction Decomposition: The integrand is $\frac{2x^2+1}{(x^2-4)(x^2-1)}$. Let $y = x^2$. Then the expression becomes $\frac{2y+1}{(y-4)(y-1)}$. We can decompose this into partial fractions: $\frac{2y+1}{(y-4)(y-1)} = \frac{A}{y-4} + \frac{B}{y-1}$ Multiply both sides by $(y-4)(y-1)$: $2y+1 = A(y-1) + B(y-4)$ To find $A$, set $y=4$: $2(4)+1 = A(4-1) \implies 9 = 3A \implies A=3$ To find $B$, set $y=1$: $2(1)+1 = B(1-4) \implies 3 = -3B \implies B=-1$ Substitute $A$ and $B$ back into the partial fraction form, and then replace $y$ with $x^2$: $\frac{2x^2+1}{(x^2-4)(x^2-1)} = \frac{3}{x^2-4} - \frac{1}{x^2-1}$

2. Integration: Now, integrate each term: $\int \frac{(2x^2+1)}{(x^2-4)(x^2-1)} dx = \int \left( \frac{3}{x^2-4} - \frac{1}{x^2-1} \right) dx$ $= 3 \int \frac{1}{x^2-2^2} dx - \int \frac{1}{x^2-1^2} dx$ Using the standard integral formula $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$: For the first term, $a=2$: $3 \int \frac{1}{x^2-2^2} dx = 3 \left( \frac{1}{2 \cdot 2} \log \left| \frac{x-2}{x+2} \right| \right) = \frac{3}{4} \log \left| \frac{x-2}{x+2} \right|$ For the second term, $a=1$: $- \int \frac{1}{x^2-1^2} dx = - \left( \frac{1}{2 \cdot 1} \log \left| \frac{x-1}{x+1} \right| \right) = - \frac{1}{2} \log \left| \frac{x-1}{x+1} \right|$ Combining these results, the integral is: $I = \frac{3}{4} \log \left| \frac{x-2}{x+2} \right| - \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + C$

3. Manipulate to Match the Given Form: The given form is $\log\left[\left(\frac{x+1}{x-1}\right)^a \left(\frac{x-2}{x+2}\right)^b \right]+c$. We need to transform our result to match this form. Using the logarithm property $\log\left(\frac{M}{N}\right) = -\log\left(\frac{N}{M}\right)$, we can rewrite the second term: $- \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| = - \frac{1}{2} \left( - \log \left| \frac{x+1}{x-1} \right| \right) = \frac{1}{2} \log \left| \frac{x+1}{x-1} \right|$ So the integral becomes: $I = \frac{3}{4} \log \left| \frac{x-2}{x+2} \right| + \frac{1}{2} \log \left| \frac{x+1}{x-1} \right| + C$ Rearranging the terms to match the order in the question: $I = \frac{1}{2} \log \left| \frac{x+1}{x-1} \right| + \frac{3}{4} \log \left| \frac{x-2}{x+2} \right| + C$ Now, use the logarithm property $k \log M = \log M^k$: $I = \log \left| \left( \frac{x+1}{x-1} \right)^{1/2} \right| + \log \left| \left( \frac{x-2}{x+2} \right)^{3/4} \right| + C$ Finally, use the logarithm property $\log M + \log N = \log (MN)$: $I = \log \left[ \left( \frac{x+1}{x-1} \right)^{1/2} \left( \frac{x-2}{x+2} \right)^{3/4} \right] + C$

4. Compare Exponents: Comparing this result with the given form $\log\left[\left(\frac{x+1}{x-1}\right)^a \left(\frac{x-2}{x+2}\right)^b \right]+c$: We can directly identify the values of 'a' and 'b': $a = \frac{1}{2}$ $b = \frac{3}{4}$