Question

$If \int\limits^3_0 3x+1/x²+9 dx=.....,Select correct option from the given options.$

(a) π/12+log(2√2)


(b) π/3+log(2√2)

(c) π/12+log √2

(d) π/6+log(2√2)

Answer 1

HELLO DEAR,

GIVEN:-
$\sf{\int\limits^3_0 (3x+1)/(x^2+9) dx}$

NOW,

$\sf{\int\limits^3_0 [3x.dx/(x^2 + 9) + dx/(x^2 + 9)]}$

in first integral
put (x^2 + 9) = t
=> 2x .dx = dt
Also, [x = 3 , t = 18] and [x = 0 , t = 9]

$\sf{3/2\int\limits^{18}_9 {dt/t} + \int\limits^3_0 {dx/(x^2 + 3^2)}}$

$\sf{3/2[log|t|]^{^{18}}_9 + 1/3[tan^{-1}{x\over3}]^{^3}_0}$

$\sf{3/2[log|2*9| - log|9|] + 1/3[tan^{-1}(3/3) - tan^{-1}(0/3)]}$

$\sf{3/2[log2 + log9 - log9] + 1/3[\pi/4 - 0]}$

$\sf{3/2log2 + 1/3\pi/4}$

$\sf{log|2^{3/2}| + \pi/12}$

$\sf{log|(\sqrt{2})^3| + \pi/12}$

$\sf{log|2\sqrt{2}| + \pi/12}$

HENCE, OPTION (A) IS CORRECT

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Given, $\int\limits^3_0{\frac{(3x+1)}{(x^2+9)}}\,dx$

now, $\int\limits^3_0{\frac{(3x+1)}{(x^2+9)}}\,dx$

= $\int\limits^3_0{\frac{3x}{x^2+9}}\,dx+\int\limits^3_0{\frac{1}{x^2+9}}\,dx$

= $\frac{3}{2}\int\limits^3_0{\frac{2x}{x^2+9}}\,dx+\int\limits^3_0{\frac{1}{x^2+3^2}}\,dx$

=$\frac{3}{2}[log(x^2+9)]^3_0+\frac{1}{3}[tan^{-1}x/3]^3_0$

= $\frac{3}{2}[log(9+9)-log9]+\frac{1}{3}[tan^{-1}3/3-tan^{-1}0]$

=$\frac{3}{2}log2+\frac{1}{3}tan^{-1}1$

= $log(2)^{3/2}+\frac{\pi}{3\times4}$

= log2√2 + π/12

hence, option (a) is correct.