Question

integrate (2x + 1)/(9 - 4x ^ 2) dx​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{I=\displaystyle\int\dfrac{2x+1}{9-4x^2}\,dx}$

$\tt{\implies\,I=\displaystyle \int\bigg[\dfrac{2x}{9-4x^2}+\dfrac{1}{9-4x^2}\bigg]\,dx}$

$\tt{\implies\,I=\displaystyle \int\dfrac{2x}{9-4x^2}\,dx+\int\dfrac{1}{9-4x^2}\,dx}$

Now, let

$\tt{I=I_{1}+I_{2}}$

$\sf{\leadsto\,\blue{SOLVING\,\,\tt{I_{1}}}}$

$\tt{I_{1}=\displaystyle\int\dfrac{2x}{9-4x^2}\,dx}$

$\tt{\green{Let\,\,9-4x^2=t}}$

$\tt{\green{\implies-8x\,dx=dt}}$

$\tt{\green{\implies2x\,dx=-\dfrac{dt}{4}}}$

So,

$\tt{I_{1}=\displaystyle\int\dfrac{1}{t}\cdot\,-\dfrac{dt}{4}}$

$\tt{\implies\,I_{1}=-\dfrac{1}{4}\displaystyle\int\dfrac{1}{t}\,dt}$

$\tt{\implies\,I_{1}=-\dfrac{1}{4}\cdot\,\ln|t|+c_{1}}$

$\boxed{\tt{\pink{\implies\,I_{1}=-\dfrac{1}{4}\,\ln|9-4x^2|+c_{1}}}}$

$\sf{\leadsto\,\blue{SOLVING\,\,\tt{I_{2}}}}$

$\tt{I_{2}=\displaystyle\int\dfrac{1}{9-4x^2}\,dx}$

We know,

$\sf{\green{\displaystyle\,\int\dfrac{dx}{(a)^2-(bx)^2}=\dfrac{1}{2ab}\ln\bigg|\dfrac{a+bx}{a-bx}\bigg|+c}}$

So,

$\tt{I_{2}=\displaystyle\int\dfrac{1}{(3)^2-(2x)^2}\,dx}$

$\tt{\implies\,I_{2}=\dfrac{1}{2\cdot3\cdot2}\,\,\ln\bigg|\dfrac{3+2x}{3-2x}\bigg|+c_{2}}$

$\boxed{\pink{\tt{\implies\,I_{2}=\dfrac{1}{12}\,\,\ln\bigg|\dfrac{3+2x}{3-2x}\bigg|+c_{2}}}}$

Now, we have,

$\tt{I=I_{1}+I_{2}}$

$\tt{\implies\,I=-\dfrac{1}{4}\,\ln|9-4x^2|+c_{1}+\dfrac{1}{12}\,\ln\bigg|\dfrac{3+2x}{3-2x}\bigg|+c_{2}}$

$\tt{\implies\,I=-\dfrac{1}{4}\,\ln|9-4x^2|+\dfrac{1}{12}\,\ln\bigg|\dfrac{3+2x}{3-2x}\bigg|+C\,\,\,\,\,\,\,\,,where\,\,C=c_{1}+c_{2}}$

$\tt{\implies\,I=-\dfrac{1}{4}\,\ln|(3-2x)(3+2x)|+\dfrac{1}{12}\,\ln\bigg|\dfrac{3+2x}{3-2x}\bigg|+C}$

$\tt{\implies\,I=-\dfrac{1}{4}\,\ln|3-2x|-\dfrac{1}{4}\,\ln|3+2x|+\dfrac{1}{12}\,\ln|3+2x|-\dfrac{1}{12}\,\ln|3-2x|+C}$

$\tt{\implies\,I=-\bigg(\dfrac{1}{4}+\dfrac{1}{12}\bigg)\,\ln|3-2x|+\bigg(\dfrac{1}{12}-\dfrac{1}{4}\bigg)\,\ln|3+2x|+C}$

$\tt{\implies\,I=-\bigg(\dfrac{3+1}{12}\bigg)\,\ln|3-2x|+\bigg(\dfrac{1-3}{12}\bigg)\,\ln|3+2x|+C}$

$\tt{\implies\,I=-\bigg(\dfrac{4}{12}\bigg)\,\ln|3-2x|+\bigg(\dfrac{-2}{12}\bigg)\,\ln|3+2x|+C}$

$\purple{\boxed{\tt{\implies\,I=-\dfrac{1}{3}\,\ln|3-2x|-\dfrac{1}{6}\,\ln|3+2x|+C}}}$