If -dxx-4-x-5-13ln- -fx-fx- -C where -C are constants- then which of the following is-are CORRECT-

Let, I=∫dx(x 4)√(x+5) Put x+5=t^2 ⇒ dx=2t dt ⇒ I=∫2t dt(t^2 9)(t) =∫2 dtt^2 9 {∵∫1t^2 a^2 dx= 12aln| t at+a|+C } nbsp; =13ln| t 3t+3|+C nbsp; =13ln| √ ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

If ∫dxx-4√x+5...

Question

If $\int \frac{d x}{(x - 4) \sqrt{x + 5}} = \frac{1}{3} ln ∣ ∣ ∣ ∣ \frac{\sqrt{f (x)} - α}{\sqrt{f (x)} + β} ∣ ∣ ∣ ∣ + C$ where $(α, β, C)$ are constants, then which of the following is/are CORRECT?

α = β

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α \neq β

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f (x) = x + 5

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f (x) = \sqrt{x + 5}

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Solution

The correct option is C $f (x) = x + 5$
Let, $I = \int \frac{d x}{(x - 4) \sqrt{x + 5}}$
Put $x + 5 = t^{2}$
$\Rightarrow d x = 2 t d t$
$\Rightarrow I = \int \frac{2 t d t}{(t^{2} - 9) (t)} = \int \frac{2 d t}{t^{2} - 9} {∵ \int \frac{1}{t^{2} - a^{2}} d x = \frac{1}{2 a} ln ∣ ∣ ∣ \frac{t - a}{t + a} ∣ ∣ ∣ + C} = \frac{1}{3} ln ∣ ∣ ∣ \frac{t - 3}{t + 3} ∣ ∣ ∣ + C = \frac{1}{3} ln ∣ ∣ ∣ \frac{\sqrt{x + 5} - 3}{\sqrt{x + 5} + 3} ∣ ∣ ∣ + C$

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If ∫dx(x−4)√x+5=13ln∣∣ ∣∣√f(x)−α√f(x)+β∣∣ ∣∣+C where (α,β,C) are constants, then which of the following is/are CORRECT?

The correct option is C f(x)=x+5Let, I=∫dx(x−4)√x+5 Put x+5=t2 ⇒dx=2t dt ⇒I=∫2t dt(t2−9)(t)=∫2 dtt2−9{∵∫1t2−a2dx=12aln∣∣∣t−at+a∣∣∣+C}=13ln∣∣∣t−3t+3∣∣∣+C=13ln∣∣∣√x+5−3√x+5+3∣∣∣+C

If $\int \frac{d x}{(x - 4) \sqrt{x + 5}} = \frac{1}{3} ln ∣ ∣ ∣ ∣ \frac{\sqrt{f (x)} - α}{\sqrt{f (x)} + β} ∣ ∣ ∣ ∣ + C$ where $(α, β, C)$ are constants, then which of the following is/are CORRECT?