The correct option is A 12√(2)tan^ 1(2x 1√(2))+C Now, ∫14x^2+4x+3 dx =∫14x^2+2.2x+1+2 dx =∫1(2x^2+1)^2+2 dx =12√(2)tan^ 12x+1√(2)+c ...

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Definite Integral as Limit of Sum

∫14x2+4x+3dx=...

Question

$\int \frac{1}{4 x^{2} + 4 x + 3} d x = ?$

\frac{1}{2 \sqrt{2}} {tan}^{- 1} (\frac{2 x - 1}{\sqrt{2}}) + C

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\frac{1}{2 \sqrt{2}} {tan}^{- 1} (\frac{2 x + 1}{\sqrt{2}}) + C

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\frac{1}{2} {tan}^{- 1} (2 x - 1) + C

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\frac{1}{2} {tan}^{- 1} (2 x + 1) + C

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Solution

The correct option is A $\frac{1}{2 \sqrt{2}} {tan}^{- 1} (\frac{2 x - 1}{\sqrt{2}}) + C$
Now,
$\int \frac{1}{4 x^{2} + 4 x + 3} d x$
$= \int \frac{1}{4 x^{2} + 2.2 x + 1 + 2} d x$
$= \int \frac{1}{(2 x^{2} + 1)^{2} + 2} d x$
$= \frac{1}{2 \sqrt{2}} {tan}^{- 1} \frac{2 x + 1}{\sqrt{2}} + c$

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Similar questions

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

2 {tan}^{- 1} x .

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

f (x) = {sin}^{- 1} (\frac{2 x}{1 + x^{2}})

f^{'} (2) = - \frac{2}{5}

{sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = π - 2 {tan}^{- 1} x

\forall x > 1

x \geq 1, t h e n 2 T a n^{- 1} x + S i n^{- 1} (\frac{2 x}{1 + x^{2}})

3 {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

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