Evaluate the integral using substitution- -20 dxx-4-x2

∫^2_0 dxx+4 x^2 = ∫^2_0 dxx^2 x 4 ∫^2_0dxx^2 x+(12)^2 (12)^2 4 = ∫^2_0dx(x 12)^2 14 4 = ∫^2_0dx(x 12)^2 174 Let (x 12)=t Differentiating with respect to ...

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Byju's Answer

Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

Evaluate the ...

Question

Evaluate the integral using substitution:
$\int_{0}^{2} \frac{d x}{x + 4 - x^{2}}$

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Solution

$\int_{0}^{2} \frac{d x}{x + 4 - x^{2}}$

$= - \int_{0}^{2} \frac{d x}{x^{2} - x - 4}$

$- \int_{0}^{2} \frac{d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 4}$

$= - \int_{0}^{2} \frac{d x}{{(x - \frac{1}{2})}^{2} - \frac{1}{4} - 4}$

$= - \int_{0}^{2} \frac{d x}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}}$

Let

$(x - \frac{1}{2}) = t$

Differentiating with respect to $t$

$\frac{d x}{d t} = 1 \Rightarrow d x = d t$

The new limits are

When $x = 0$

$\Rightarrow t = 0 - \frac{1}{2} = - \frac{1}{2}$

When $x = 2$

$\Rightarrow t = 2 - \frac{1}{2} = - \frac{3}{2}$

Now,

$\int_{0}^{2} \frac{d x}{x + 4 - x^{2}}$

$= - \int_{0}^{2} \frac{d x}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}}$

$= - \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{t^{2} - \frac{17}{4}}$

$= - \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{t^{2} - {(\frac{\sqrt{17}}{2})}^{2}}$

$= - {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1}{2 \times \frac{\sqrt{17}}{2}} \times log ∣ ∣ ∣ ∣ ∣ ∣ \frac{t - \frac{\sqrt{17}}{2}}{t + \frac{\sqrt{17}}{2}} ∣ ∣ ∣ ∣ ∣ ∣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}_{- \frac{1}{2}}^{\frac{3}{2}}$

$[∵ \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} log ∣ ∣ ∣ \frac{x - a}{x + a} ∣ ∣ ∣ + C]$

$= - \frac{1}{\sqrt{17}} {[log ∣ ∣ ∣ \frac{2 t - \sqrt{17}}{2 t + \sqrt{17}} ∣ ∣ ∣]}_{- \frac{1}{2}}^{\frac{3}{2}}$

$= - \frac{1}{\sqrt{17}} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎛ ⎜ ⎜ ⎜ ⎝ log ∣ ∣ ∣ ∣ ∣ ∣ \frac{2 ⎛ ⎝ \frac{3}{2} ⎞ ⎠ - \sqrt{17}}{2 ⎛ ⎝ \frac{3}{2} ⎞ ⎠ + \sqrt{17}} ∣ ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎟ ⎠ - ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ log ∣ ∣ ∣ ∣ ∣ ∣ \frac{2 (\frac{- 1}{2}) - \sqrt{17}}{2 (\frac{- 1}{2}) + \sqrt{17}} ∣ ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{3 - \sqrt{17}}{3 + \sqrt{17}} ∣ ∣ ∣) - (log ∣ ∣ ∣ \frac{- 1 - \sqrt{17}}{- 1 + \sqrt{17}} ∣ ∣ ∣) -]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{3 - \sqrt{17}}{3 + \sqrt{17}} ∣ ∣ ∣) - (log ∣ ∣ ∣ \frac{\sqrt{17} + 1}{\sqrt{17} - 1} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{3 - \sqrt{17}}{3 + \sqrt{17}} ∣ ∣ ∣) + (log ∣ ∣ ∣ \frac{\sqrt{17} - 1}{\sqrt{17} + 1} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ ∣ \frac{3 - \sqrt{17}}{3 + \sqrt{17}} \times (\frac{\sqrt{17} - 1}{\sqrt{17} + 1}) ∣ ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{3 \sqrt{17} - 3 - 17 + \sqrt{17}}{3 \sqrt{17} + 3 + 17 + \sqrt{17}} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{4 \sqrt{17} - 20}{4 \sqrt{17} + 20} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{\sqrt{17} - 5}{\sqrt{17} + 5} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{\sqrt{17} - 5}{\sqrt{17} + 5} \times (\frac{\sqrt{17} + 5}{\sqrt{17} + 5}) ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} ⎡ ⎢ ⎢ ⎢ ⎣ ⎛ ⎜ ⎜ ⎝ log ∣ ∣ ∣ ∣ ∣ \frac{{(\sqrt{17})}^{2} - 5^{2}}{(\sqrt{17} + 5^{2})} ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{17 - 25}{17 + 25 + 10 \sqrt{17}} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{- 8}{42 + 10 \sqrt{17}} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [(log ∣ ∣ ∣ \frac{- 4}{21 + 5 \sqrt{17}} ∣ ∣ ∣)]$

$= - \frac{1}{\sqrt{17}} [log (\frac{4}{21 + 5 \sqrt{17}})]$

$= - \frac{1}{\sqrt{17}} [log (\frac{21 + 5 \sqrt{17}}{4})]$

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Evaluate the integral using substitution: ∫20dxx+4−x2

Evaluate the integral using substitution:
$\int_{0}^{2} \frac{d x}{x + 4 - x^{2}}$