If -dxx2-1-x-p-ln-fx-1fx-1-q-tan-1gx-C-Then the value of f4p-q- g1-Where p-q are fixed constants and C is integration constant

Let I=∫dx(x^2 1)√(x) Put x=t^2, dx=2tdt The integral becomes : I=∫2dtt^4 1=∫[ 1t^2 1 1t^2+1]dt =12ln|t 1t+1| tan^ 1t+C =12ln|√(x) 1√(x)+1| tan^ 1√(x)+C ⇒ ...

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Standard XII

Physics

Basic Differentiation Rule

If ∫dxx2-1√x=...

Question

If $\int \frac{d x}{(x^{2} - 1) \sqrt{x}} = p \cdot ln ∣ ∣ ∣ \frac{f (x) - 1}{f (x) + 1} ∣ ∣ ∣ + q \cdot {tan}^{- 1} (g (x)) + C .$
Then the value of $\frac{f (4)}{p} - q \cdot g (1) =$
(Where $p, q$ are fixed constants and $C$ is integration constant)

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Solution

Let $I = \int \frac{d x}{(x^{2} - 1) \sqrt{x}}$
Put $x = t^{2}, d x = 2 t d t$
The integral becomes :
$I = \int \frac{2 d t}{t^{4} - 1} = \int [\frac{1}{t^{2} - 1} - \frac{1}{t^{2} + 1}] d t = \frac{1}{2} ln ∣ ∣ ∣ \frac{t - 1}{t + 1} ∣ ∣ ∣ - {tan}^{- 1} t + C = \frac{1}{2} ln ∣ ∣ ∣ \frac{\sqrt{x} - 1}{\sqrt{x} + 1} ∣ ∣ ∣ - {tan}^{- 1} \sqrt{x} + C$
$\Rightarrow p = \frac{1}{2}, q = - 1, f (x) = \sqrt{x}, g (x) = \sqrt{x}$
So, $\frac{f (4)}{p} - q \cdot g (1) = 5$

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Basic Differentiation Rule

Standard XII Physics

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If ∫dx(x2−1)√x=p⋅ln∣∣∣f(x)−1f(x)+1∣∣∣+q⋅tan−1(g(x))+C. Then the value of f(4)p−q⋅g(1)= (Where p,q are fixed constants and C is integration constant)

Let I=∫dx(x2−1)√x Put x=t2,dx=2tdt The integral becomes : I=∫2dtt4−1=∫[1t2−1−1t2+1]dt =12ln∣∣∣t−1t+1∣∣∣−tan−1t+C =12ln∣∣∣√x−1√x+1∣∣∣−tan−1√x+C ⇒p=12,q=−1,f(x)=√x,g(x)=√x So, f(4)p−q⋅g(1)=5

If $\int \frac{d x}{(x^{2} - 1) \sqrt{x}} = p \cdot ln ∣ ∣ ∣ \frac{f (x) - 1}{f (x) + 1} ∣ ∣ ∣ + q \cdot {tan}^{- 1} (g (x)) + C .$
Then the value of $\frac{f (4)}{p} - q \cdot g (1) =$
(Where $p, q$ are fixed constants and $C$ is integration constant)