Mathematics

$$\int {\frac{{2x - 1}}{{\sqrt {9{x^2} - 4} }}dx} $$


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{2x-1}{\sqrt{9{{x}^{2}}-4}}}dx$$

$$I=\int{\dfrac{2x}{\sqrt{9{{x}^{2}}-4}}}dx-\int{\dfrac{1}{\sqrt{9{{x}^{2}}-4}}}dx$$

$$I={{I}_{1}}-{{I}_{2}}$$      …….. (1)

So,

$${{I}_{1}}=\int{\dfrac{2x}{\sqrt{9{{x}^{2}}-4}}}dx$$

Let $$t=9{{x}^{2}}$$

$$ \dfrac{dt}{dx}=9\times 2x $$

$$ \dfrac{dt}{9}=2xdx $$

Therefore,

$$ {{I}_{1}}=\dfrac{1}{9}\int{\dfrac{1}{\sqrt{t-4}}}dt $$

$$ {{I}_{1}}=\dfrac{1}{9}\left[ 2\sqrt{t-4} \right]+C $$

On putting the value of t, we get

$${{I}_{1}}=\dfrac{2}{9}\left[ \sqrt{9{{x}^{2}}-4} \right]+C$$

Now,

$$ {{I}_{2}}=\int{\dfrac{1}{\sqrt{9{{x}^{2}}-4}}}dx $$

$$ {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{\left( \dfrac{2}{3} \right)}^{2}}}}}dx $$

We know that

$$\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}=\log \left[ x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right]+C$$

Therefore,

$$ {{I}_{2}}=\dfrac{1}{3}\left[ \log \left( x+\sqrt{{{x}^{2}}-{{\left( \dfrac{2}{3} \right)}^{2}}} \right) \right]+C $$

$$ {{I}_{2}}=\dfrac{1}{3}\left[ \log \left( x+\sqrt{{{x}^{2}}-\dfrac{4}{9}} \right) \right]+C $$

From equation (1), we get

$$I=\dfrac{2}{9}\left[ \sqrt{9{{x}^{2}}-4} \right]-\dfrac{1}{3}\left[ \log \left( x+\sqrt{{{x}^{2}}-\dfrac{4}{9}} \right) \right]+C$$

Hence, the value is $$\dfrac{2}{9}\left[ \sqrt{9{{x}^{2}}-4} \right]-\dfrac{1}{3}\left[ \log \left( x+\sqrt{{{x}^{2}}-\dfrac{4}{9}} \right) \right]+C$$.

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