Calculust 7

Notice that $\frac{d}{dx} \left(x - \frac{1}{x}\right) = 1 + \frac{1}{x^2}$ After dividing the numerator and denominator of the integrand by $x^2$ , we can rewrite the integral as, $\large \int_{1}^{\frac{1+\sqrt{5}}{2}} \dfrac{1 + \frac{1}{x^2}}{x^2-1+ \frac{1}{x^2}} \ln\left( 1+x-\dfrac{1}{x} \right) \ dx$ So if $t = x - \frac{1}{x}$ , $dt = \left(1 + \frac{1}{x^2}\right) dx$ $t^2 + 1 = x^2-1+ \frac{1}{x^2}$ After computing the new limits, the integral becomes $\int_{0}^{1} \dfrac{\ln\left( 1 + t \right)}{1 + t^2} dt$ The $1 + t^2$ should remind you of the trig identity $1 + \tan^2 \theta = \sec^2 \theta$ and we also have, $\frac{d}{d\theta} \tan \theta = \sec^2 \theta$ So we substitute $t = \tan \theta$ and we now have $\int_{0}^{\frac{\pi}{4}} \dfrac{\ln(1 + \tan \theta) \sec^2 \theta d\theta}{\sec^2 \theta} = \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan \theta) d\theta$ We now apply a useful (but simple to prove) property of definite integrals - $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx = \frac{1}{2} \int_{a}^{b} f(x) + f(a + b - x) dx$ And so, we evaluate $\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan \theta) + \ln\left(1 + \tan \left(\frac{\pi}{4} - \theta\right)\right)d\theta$ Now, $\tan \left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta}$ and $\ln u + \ln v = \ln uv$ Using these facts, our integral is, $\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan \theta + 1 - \tan \theta) d\theta= \frac{1}{2} \large \int_{0}^{\frac{\pi}{4}} \ln 2 \cdot d\theta$ As the integrand is constant, we can integrate easily and get $I = \frac{\ln 2}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8} \ln 2$ So finally, $a + b = 8 + 2 = 10$ .

$Hints:$

Firstly use $t = x - \frac{1}{x}$ ,

U will get

$\int_{0}^{1} \dfrac{\ln\left( 1 + t \right)}{1 + t^2} dt$

then just take

$I(α)=\int_{0}^{1} \dfrac{\ln\left( 1 + αt \right)}{1 + t^2} dt$

By differentiating this expression w.r.t $α$ we get,

$I'(α)=\int_{0}^{1} \dfrac{t}{(1+αt)(1 + t^2)} dt$

Breaking into partial fractions then again integrating w.r.t $α$ , u will get $I=\frac{π}{4}ln(2)-I$

we can easily get
$\frac{π}{8} ln(2)$ .