Question

solve it question please friend or teacher
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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\int\dfrac{\sqrt{2x+3}}{x}\,dx}$

$\bf{Put\,\,\,2x+3={t}^{2}}$

$\bf{\mapsto\,2\,dx=2{t}\,dt}$

$\bf{\mapsto\,dx={t}\,dt}$

So,

$\displaystyle\rm{\int\dfrac{\sqrt{{t}^{2}}}{\dfrac{{t}^{2}-3}{2}}\cdot\,t\,dt}$

$\displaystyle\rm{=\int\dfrac{2\,t}{{t}^{2}-3}\cdot\,t\,dt}$

$\displaystyle\rm{=\int\dfrac{2\,{t}^{2}}{{t}^{2}-3}\,dt}$

$\displaystyle\rm{=\int\dfrac{2\,{t}^{2}-6+6}{{t}^{2}-3}\,dt}$

$\displaystyle\rm{=\int\dfrac{2\,{t}^{2}-6}{{t}^{2}-3}\,dt+\int\dfrac{6}{{t}^{2}-3}\,dt}$

$\displaystyle\rm{=\int\dfrac{2\left({t}^{2}-3\right)}{{t}^{2}-3}\,dt+6\int\dfrac{dt}{{t}^{2}-3}}$

$\displaystyle\rm{=\int2\,dt+6\int\dfrac{dt}{\left(t\right)^{2}-\left(\sqrt{3}\right)^2}}$

We know,

$\boxed{\displaystyle\blue{\bf{\int\dfrac{dx}{{x}^{2}-{a}^{2}}=\dfrac{1}{2a}\cdot\ln\left|\dfrac{x-a}{x+a}\right|+C}}}$

So,

$\displaystyle\rm{=2t+6\cdot\dfrac{1}{2\sqrt{3}}\ln\left|\dfrac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C}$

$\displaystyle\rm{=2t+\sqrt{3}\ln\left|\dfrac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C}$

Put the value of t,

$\displaystyle\rm{=2\sqrt{2x+3}+\sqrt{3}\ln\left|\dfrac{\sqrt{2x+3}-\sqrt{3}}{\sqrt{2x+3}+\sqrt{3}}\right|+C}$