Question

integrate this plz it's urgent​

Answer 1

Let (3x - 4) = y

Differentiating :-

$\sf{\dfrac{dy}{dx} =\dfrac{d(3x-4)}{dx}}$

$\sf{=3.\dfrac{d}{dx}[x]+\dfrac{d}{dx}[-4]}$

The constant term is moved out.

$\sf{=3\times1+0}$

The derivative of differentiation variable is 1, and the derivative of a constant term is 0.

$\sf{=3}$

So, $\boxed{\sf{\dfrac{dy}{dx }=3 }}$

$\sf{\implies dx=\dfrac{1}{3}dy }$

$\sf{=\dfrac{1}{3}\int\limits {y^\frac{3}{2}} \, dy }$

$\rule{100}{1.5}$

$\sf{Now,\ solving:-}\\\\\sf{=\int y^\frac{3}{2}dy}\\\\\sf{=\dfrac{y^\frac{3}{2}^+^1 }{\dfrac{3}{2}+1} }$

$\sf{=\dfrac{y^\frac{5}{2}}{\dfrac{5}{2}} }\\\\\\\sf{=\dfrac{2y^\frac{5}{2}}{5} }$

$\rule{100}{1.5}$

$\sf{Now,\ \dfrac{1}{3}\int\limits {y^\frac{3}{2}} \, dy }$

$\sf{=\dfrac{1}{3}\times \dfrac{2y^\frac{5}{2}}{5}+C}$

$\sf{=\dfrac{2y^\frac{5}{2}}{15}+C }$

Putting the value of y as 3x - 4 :-

$\large\boxed{\boxed{\sf{=\dfrac{2(3x-4)^\frac{5}{2}}{15}+C}}}$                    ------ANSWER

$\rule{200}{2}$

• A constant C is always added after integration.

Answer 2

Explanation:

your ans is in above attachement....

thnks for asking sir...mark me as brainellist if u got the ans...