Question

please
$\red {\tt{integrate \frac{ (3x + 1)}{(x ^ 2 + 9)} dx \: from \: 0 \: to \: 3}}$
​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\int\limits_0^3\:\left(\:\dfrac{3x\:+\:1}{x^2\:+\:9}\:\right)\:dx\:=\:\ln\:(\:2\:\sqrt{2}\:)\:+\:\dfrac{\pi}{12}}}}$

Step-by-step-explanation:

The given definite integral is

$\displaystyle{\sf\:I\:=\:\int\limits_0^3\:\left(\:\dfrac{3x\:+\:1}{x^2\:+\:9}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I\:=\:\int\limits_0^3\:\left(\:\dfrac{3x}{x^2\:+\:9}\:\right)\:dx\:+\:\int\limits_0^3\:\left(\:\dfrac{1}{x^2\:+\:9}\:\right)\:dx}$

Let,

$\displaystyle{\sf\:\int\limits_0^3\:\left(\:\dfrac{3x}{x^2\:+\:9}\:\right)\:dx\:=\:I_1}$

And

$\displaystyle{\sf\:\int\limits_0^3\:\left(\:\dfrac{1}{x^2\:+\:9}\:\right)\:dx\:=\:I_2}$

Now,

$\displaystyle{\sf\:I_1\:=\:\int\limits_0^3\:\left(\:\dfrac{3x}{x^2\:+\:9}\:\right)\:dx}$

Put x² + 9 = t

Differentiating both sides w.r.t. x, we get,

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:x^2\:+\:9\:)\:=\:\dfrac{d}{dx}\:t}$

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:x^2\:)\:+\:\dfrac{d}{dx}\:9\:=\:\dfrac{dt}{dx}}$

$\displaystyle{\implies\sf\:2x\:+\:0\:=\:\dfrac{dt}{dx}}$

$\displaystyle{\implies\sf\:dx\:=\:\dfrac{dt}{2x}}$

For lower limit, x = 0,

x² + 9 = t

⇒ 0 + 9 = t

⇒ t = 9

For upper limit, x = 3,

x² + 9 = t

⇒ ( 3 )² + 9 = t

⇒ t = 9 + 9

⇒ t = 18

Now, using this,

$\displaystyle{\implies\sf\:I_1\:=\:\int\limits_9^{18}\:\left(\:\dfrac{3\:\cancel{x}}{t}\:\dfrac{1}{2\:\cancel{x}}\:\right)\:dt}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:\int\limits_9^{18}\:\dfrac{1}{t}\:dt}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:\ln\:(\:t\:)\:\bigg|_9^{18}}$

By re-substituting t = x² + 9,

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:\ln\:(\:x^2\:+\:9\:)\:\bigg|_0^3}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:[\:\ln\:(\:3^2\:+\:9\:)\:-\:\ln\:(\:0^2\:+\:9\:)\:]}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:[\:\ln\:(\:9\:+\:9\:)\:-\:\ln\:(\:0\:+\:9\:)\:]}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:[\:\ln\:(\:18\:)\:-\:\ln\:(\:9\:)\:]}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:\times\:\ln\:\left(\:\dfrac{18}{9}\:\right)}$

$\displaystyle{\implies\sf\:I_1\:=\:\dfrac{3}{2}\:\times\:\ln\:(\:2\:)}$

$\displaystyle{\implies\sf\:I_1\:=\:\ln\:(\:2^{\frac{3}{2}}\:)}$

$\displaystyle{\implies\sf\:I_1\:=\:\ln\:[\:(\:2^{\frac{1}{2}} \:)^3\:]}$

$\displaystyle{\implies\sf\:I_1\:=\:\ln\:[\:(\:\sqrt{2}\:)^3\:]}$

$\displaystyle{\implies\sf\:I_1\:=\:\ln\:(\:2\:\sqrt{2}\:)}$

Now,

$\displaystyle{\sf\:I_2\:=\:\int\limits_0^3\:\left(\:\dfrac{1}{x^2\:+\:9}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I_2\:=\:\int\limits_0^3\:\left(\:\dfrac{1}{x^2\:+\:3^2}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I_2\:=\:\int\limits_0^3\:\left(\:\dfrac{1}{1\:+\:\left(\:\dfrac{x}{3}\:\right)^2}\:\dfrac{dx}{3^2}\:\right)}$

Put $\displaystyle{\sf\:\dfrac{x}{3}\:=\:t}$

Differentiating both sides w.r.t. x, we get,

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:\left(\:\dfrac{x}{3}\:\right)\:=\:\dfrac{dt}{dx}}$

$\displaystyle{\implies\sf\:\dfrac{1}{3}\:\dfrac{d}{dx}\:x\:=\:\dfrac{dt}{dx}}$

$\displaystyle{\implies\sf\:\dfrac{1}{3}\:=\:\dfrac{dt}{dx}}$

$\displaystyle{\implies\sf\:dx\:=\:3\:dt}$

For lower limit, x = 0,

$\displaystyle{\sf\:\dfrac{x}{3}\:=\:t}$

$\displaystyle{\implies\sf\:\dfrac{0}{3}\:=\:t}$

$\displaystyle{\implies\boxed{\sf\:t\:=\:0}}$

For upper limit, x = 3,

$\displaystyle{\sf\:\dfrac{x}{3}\:=\:t}$

$\displaystyle{\implies\sf\:\dfrac{3}{3}\:=\:t}$

$\displaystyle{\implies\boxed{\sf\:t\:=\:1}}$

Now, using this,

$\displaystyle{\implies\sf\:I_2\:=\:\int\limits_0^1\:\left(\:\dfrac{1}{1\:+\:t^2}\:\right)\:\dfrac{3\:dt}{3^2}}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\int\limits_0^1\:\left(\:\dfrac{1}{1\:+\:t^2}\:\right)\:dt}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\tan^{-\:1}\:(\:t\:)\:\bigg|_0^1}$

By re-substituting $\displaystyle{\sf\:t\:=\:\dfrac{x}{3}}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\tan^{-\:1}\:\left(\:\dfrac{x}{3}\:\right)\:\bigg|_0^3}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\left[\:\tan^{-\:1}\:\left(\:\dfrac{3}{3}\:\right)\:-\:\tan^{-\:1}\:\left(\:\dfrac{0}{3}\:\right)\:\right]}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\left(\:\tan^{-\:1}\:(\:1\:)\:-\:\tan^{-\:1}\:(\:0\:)\:\right)}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{1}{3}\:\times\:\dfrac{\pi}{4}}$

$\displaystyle{\implies\sf\:I_2\:=\:\dfrac{\pi}{12}}$

Now,

$\displaystyle{\sf\:I\:=\:I_1\:+\:I_2}$

$\displaystyle{\implies\sf\:I\:=\:\ln\:(\:2\:\sqrt{2}\:)\:+\:\dfrac{\pi}{12}}$

$\displaystyle{\therefore\underline{\boxed{\red{\sf\:\int\limits_0^3\:\left(\:\dfrac{3x\:+\:1}{x^2\:+\:9}\:\right)\:dx\:=\:\ln\:(\:2\:\sqrt{2}\:)\:+\:\dfrac{\pi}{12}}}}}$

Answer 2

Answer:

Step-by-step explanation:

Hope it helps you