Visit the website manishverma.site for latest posts, courses, admission & more.

### $f\left( {\frac{{2x - 1}}{{x - 3}}} \right) = {x^2}$, $\int {f(x)dx = ?}$

Let, $\frac{{2x - 1}}{{x - 3}} = t$

$\therefore x = \frac{{3t - 1}}{{t - 2}}$

$\therefore f(t) = {\left( {\frac{{3t - 1}}{{t - 2}}} \right)^2}$

Now, $\int {f(t)dt} = \int {{{\left( {\frac{{3t - 1}}{{t - 2}}} \right)}^2}dt}$

$= \int {{{\left[ {\frac{{3(t - 2) + 5}}{{t - 2}}} \right]}^2}dt}$

$= \int {{{\left( {3 + \frac{5}{{t - 2}}} \right)}^2}dt}$

$= \int {9 + \frac{{30}}{{t - 2}} + \frac{{25}}{{{{(t - 2)}^2}}}dt}$

$= 9t + 30\ln |t - 2| - \frac{{25}}{{t - 2}} + C$

The integral asked $=9x + 30\ln |x - 2| - \frac{{25}}{{x - 2}} + C$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$