4int_0^1left(frac1sqrt3 + x^2 + sqrt1 + x^2right)mathrmdx - 3log_eleft(sqrt3right)$4\int_{0}^{1}\left(\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right)\mathrm{d}x - 3\log_{e}\left(\sqrt{3}\right)$ is equal to:

### Related Formula int sqrta^2 + x^2 dx = fracx2 sqrta^2 + x^2 + fraca^22 lnleft| x + sqrta^2 + x^2 right|$$\int \sqrt{a^2 + x^2} dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln\left| x + \sqrt{a^2 + x^2} \right|$$ ### Core Logic We first rationalise the denominator to split the integral into two standard integration terms. ### Step 1: Rationalise the integrand Multiply the numerator and denominator by sqrt3+x^2 - sqrt1+x^2$\sqrt{3+x^2} - \sqrt{1+x^2}$: frac1sqrt3 + x^2 + sqrt1 + x^2 = fracsqrt3 + x^2 - sqrt1 + x^2(3+x^2) - (1+x^2) = fracsqrt3 + x^2 - sqrt1 + x^22$$\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3+x^2) - (1+x^2)} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2}$$ Thus, the integral expression simplifies to: I = 4 int_0^1 left( fracsqrt3 + x^2 - sqrt1 + x^22 right) dx - 3log_eleft(sqrt3right)$$I = 4 \int_{0}^{1} \left( \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \right) dx - 3\log_{e}\left(\sqrt{3}\right)$$ I = 2 int_0^1 sqrt3 + x^2 dx - 2 int_0^1 sqrt1 + x^2 dx - frac32 log_e 3$$I = 2 \int_{0}^{1} \sqrt{3 + x^2} dx - 2 \int_{0}^{1} \sqrt{1 + x^2} dx - \frac{3}{2} \log_{e} 3$$ ### Step 2: Evaluate the integrals For the first integral: 2 int_0^1 sqrt3 + x^2 dx = 2 left[ fracx2 sqrt3 + x^2 + frac32 lnleft| x + sqrt3 + x^2 right| right]_0^1$$2 \int_{0}^{1} \sqrt{3 + x^2} dx = 2 \left[ \frac{x}{2} \sqrt{3 + x^2} + \frac{3}{2} \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$$ = left[ x sqrt3 + x^2 + 3 lnleft| x + sqrt3 + x^2 right| right]_0^1$$= \left[ x \sqrt{3 + x^2} + 3 \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$$ = left( sqrt4 + 3 ln(1 + sqrt4) right) - left( 0 + 3 lnsqrt3 right)$$= \left( \sqrt{4} + 3 \ln(1 + \sqrt{4}) \right) - \left( 0 + 3 \ln\sqrt{3} \right)$$ = 2 + 3 ln 3 - frac32 ln 3 = 2 + frac32 ln 3$$= 2 + 3 \ln 3 - \frac{3}{2} \ln 3 = 2 + \frac{3}{2} \ln 3$$ For the second integral: -2 int_0^1 sqrt1 + x^2 dx = -2 left[ fracx2 sqrt1 + x^2 + frac12 lnleft| x + sqrt1 + x^2 right| right]_0^1$$-2 \int_{0}^{1} \sqrt{1 + x^2} dx = -2 \left[ \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$$ = - left[ x sqrt1 + x^2 + lnleft| x + sqrt1 + x^2 right| right]_0^1$$= - \left[ x \sqrt{1 + x^2} + \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$$ = - left( sqrt2 + ln(1 + sqrt2) right)$$= - \left( \sqrt{2} + \ln(1 + \sqrt{2}) \right)$$ ### Step 3: Sum the terms Now compile all terms: I = left( 2 + frac32 ln 3 right) - sqrt2 - ln(1 + sqrt2) - frac32 ln 3$$I = \left( 2 + \frac{3}{2} \ln 3 \right) - \sqrt{2} - \ln(1 + \sqrt{2}) - \frac{3}{2} \ln 3$$ I = 2 - sqrt2 - ln(1 + sqrt2)$$I = 2 - \sqrt{2} - \ln(1 + \sqrt{2})$$ ### Pattern Recognition Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus