-dxx-1-2x-3 equals to where C is integration constant

Let I=∫dx(x+1)√(2x 3) Putting 2x 3=t^2 ⇒ dx=t dt we get, I=∫t dt ( t^2+32+1 )t=∫2dtt^2+5 =2√(5)tan^ 1 ( t√(5) )+C {∵∫1t^2+a^2 dx= 1atan^ 1( ta)+C } =2√(5)tan^ ...

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Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

∫dxx+1√2x-3 e...

Question

$\int \frac{d x}{(x + 1) \sqrt{2 x - 3}}$ equals to (where $C$ is integration constant)

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Solution

Let $I = \int \frac{d x}{(x + 1) \sqrt{2 x - 3}}$
Putting $2 x - 3 = t^{2} \Rightarrow d x = t d t$
we get,
$I = \int \frac{t d t}{(\frac{t^{2} + 3}{2} + 1) t} = \int \frac{2 d t}{t^{2} + 5}$
$= \frac{2}{\sqrt{5}} {tan}^{- 1} (\frac{t}{\sqrt{5}}) + C {∵ \int \frac{1}{t^{2} + a^{2}} d x = \frac{1}{a} {tan}^{- 1} (\frac{t}{a}) + C}$
$= \frac{2}{\sqrt{5}} {tan}^{- 1} \sqrt{\frac{2 x - 3}{5}} + C$

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∫dx(x+1)√2x−3 equals to (where C is integration constant)

Let I=∫dx(x+1)√2x−3 Putting 2x−3=t2⇒dx=tdt we get, I=∫tdt(t2+32+1)t=∫2dtt2+5 =2√5tan−1(t√5)+C{∵∫1t2+a2dx=1atan−1(ta)+C} =2√5tan−1√2x−35+C

$\int \frac{d x}{(x + 1) \sqrt{2 x - 3}}$ equals to (where $C$ is integration constant)