The correct option is D 1/8log ( 4x^2 4x 3 )+1/16log2x 3/2x+1. Let I=∫ 1 x / 4 x ^ 2 4x 3 dx =∫( 1 / 2( 4 x ^ 2 4x 3 ) nbsp; 8x 4 / 8( 4 ...

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Functions

∫1-x/4x2-4x-3...

Question

$\int \frac{1 - x}{4 x^{2} - 4 x - 3} d x$

\frac{1}{8} log (4 x^{2} - 4 x - 3) + \frac{1}{16} log \frac{2 x - 3}{2 x + 1} .

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- \frac{1}{8} log (4 x^{2} - 4 x - 3) + \frac{1}{8} log \frac{2 x - 3}{2 x + 1} .

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\frac{1}{16} log (4 x^{2} - 4 x - 3) - \frac{1}{8} log \frac{2 x - 3}{2 x + 1} .

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- \frac{1}{8} log (4 x^{2} - 4 x - 3) + \frac{1}{16} log \frac{2 x - 3}{2 x + 1} .

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Solution

The correct option is D $- \frac{1}{8} log (4 x^{2} - 4 x - 3) + \frac{1}{16} log \frac{2 x - 3}{2 x + 1} .$
Let $I = \int \frac{1 - x}{4 x^{2} - 4 x - 3} d x = \int (\frac{1}{2 (4 x^{2} - 4 x - 3)} - \frac{8 x - 4}{8 (4 x^{2} - 4 x - 3)}) d x$
$\Rightarrow I = I_{1} + I_{2}$
Where
$I_{1} = - \frac{1}{8} \int \frac{8 x - 4}{4 x^{2} - 4 x - 3} d x$
Put $4 x^{2} - 4 x - 3 = t \Rightarrow (8 x - 4) d x = d t$
Therefore $I_{1} = - \frac{1}{8} log t = - \frac{1}{8} log (4 x^{2} - 4 x - 3)$
And $I_{2} = \frac{1}{2} \int \frac{1}{4 x^{2} - 4 x - 3} d x = \frac{1}{2} \int \frac{1}{{(2 x - 1)}^{2} - 4} d x$
Put $u = 2 x - 1 \Rightarrow d u =\Rightarrow 2 d x$
Therefore
$I_{2} = \frac{1}{4} \int \frac{1}{u^{2} - 4} d u = - \frac{1}{4} \int \frac{1}{4 (1 - \frac{u^{2}}{4})} d u = - \frac{1}{16} \int \frac{1}{1 - \frac{u^{2}}{4}} d u$
Now put $\frac{u}{2} = s \Rightarrow \frac{1}{2} d u = d s$
$I_{2} = - \frac{1}{8} \int \frac{1}{1 - s^{2}} d s = - \frac{1}{16} log (\frac{1 + s}{1 - s})$
$= - \frac{1}{16} log (\frac{2 + u}{2 - u}) = - \frac{1}{16} log (\frac{2 + 2 x - 1}{2 - 2 x + 1}) = \frac{1}{16} log (\frac{2 x - 3}{2 x + 1})$
Hence, option 'D' is correct.

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Similar questions

\frac{4 x - 3}{2 x + 1} - 10 (\frac{2 x + 1}{4 x - 3}) = 3

\frac{4 x - 3}{2 x + 1} - 10 (\frac{2 x + 1}{4 x - 3}) = 3, x \neq \frac{- 1}{2}, \frac{3}{4}

\frac{4 x - 3}{2 x + 1} - 10 (\frac{2 x + 1}{4 x - 3}) = 3; x \neq - \frac{1}{2}, \frac{3}{4}

Solve the equations:

(i) 5x = 3x + 24;

(ii) 8t + 5 = 2t − 31;

(iii) 7x − 10 = 4x + 11;

(iv) 4z + 3 = 6 + 2z;

(v) 2x − 1 = 14 − x;

(vi) 6x + 1 = 3(x − 1) + 7;

(vii) ;

(viii) ;

(ix) 3(x + 1) = 12 + 4 (x − 1);

(x) 2x − 5 = 3(x − 5);

(xi) 6(1 − 4x) + 7(2 + 5x) = 53;

(xii) 3(x + 6) + 2 (x + 3) = 64;

(xiii) ;

(xiv) .

4 x^{2} + 8 x + 3

(x + 1) (x + 3)

(2 x + 1) (2 x + 3)

(2 x + 2) (2 x + 5)

(2 x - 1) (2 x - 3)

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