Definite Integration
Integration
Grade Class 12
Question:
∫ \frac{x^2(1-\ln x)}{\ln^4 x - x^4} dx equals
\frac{1}{2} \ln \left| \frac{x}{\ln x} \right| - \frac{1}{4} \ln \left| \ln^2 x - x^2 \right| + C
\frac{1}{4} \ln \left| \frac{\ln x - x}{\ln x + x} \right| - \frac{1}{2} \tan^{-1} \left( \frac{\ln x}{x} \right) + C
\frac{1}{4} \ln \left| \frac{\ln x + x}{\ln x - x} \right| + \frac{1}{2} \tan^{-1} \left( \frac{\ln x}{x} \right) + C
\frac{1}{4} \ln \left| \frac{\ln x - x}{\ln x + x} \right| + \tan^{-1} \left( \frac{\ln x}{x} \right) + C
Step-by-Step Solution
Key Concept: The integral can be solved by substituting u = lnx/x, which simplifies the expression into a standard form involving partial fractions or standard integral formulas.
Let I = \int (x^2(1-lnx)) / (ln^4x - x^4) dx. Divide numerator and denominator by x^4: I = \int (1/x^2 * (1-lnx)/x) / ((lnx/x)^4 - 1) dx. Let u = lnx/x, then du = (x(1/x) - lnx)/x^2 dx = (1-lnx)/x^2 dx. The integral becomes \int du / (u^4 - 1). This is \int du / ((u^2-1)(u^2+1)) = 1/2 \int (1/(u^2-1) - 1/(u^2+1)) du = 1/2 * (1/2 ln|(u-1)/(u+1)| - tan^-1(u)) + C = 1/4 ln|(lnx/x - 1)/(lnx/x + 1)| - 1/2 tan^-1(lnx/x) + C = 1/4 ln|(lnx-x)/(lnx+x)| - 1/2 tan^-1(lnx/x) + C.
Correct Answer: 2