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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{1}{4x^2-6x+7}$$dx$

$(a)\;\frac{1}{\sqrt {17}}\tan ^{-1} \bigg\{ \frac{(x+3/2)}{\sqrt {17}/4}\bigg\}+c \\(b)\; \frac{-1}{\sqrt {17}}\tan ^{-1} \bigg\{ \frac{(x+3/2)}{\sqrt {17}/4}\bigg\}+c \\(c)\; \frac{1}{\sqrt {17}}\tan ^{-1} \bigg\{ \frac{(x-3/2)}{\sqrt {17}/4}\bigg\}+c \\ (d)\;None$
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1 Answer

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$\int \large\frac{1}{4x^2-6x+7}$$dx$
$\int \large\frac{1}{4(x^2-\Large\frac{3}{2}x+\frac{7}{4})}$$dx$
$\int \large\frac{1}{4(x^2-\Large\frac{3}{2}x+\frac{7}{4}+\frac{9}{16}-\frac{9}{16})}$$dx$
$\int \large\frac{1}{4(x^2-\Large\frac{3}{2}x+\frac{9}{16}+(\frac{\sqrt {17}}{4})^2)}$$dx$
$\large\frac{1}{4} \int \large\frac{1}{(x-\Large\frac{3}{2})^2+(\frac{\sqrt {17}}{4})^2}$$dx$
=> $\large\frac{1}{4} \times \frac{1}{\sqrt {17}}$$ \times 4 \times \tan ^{-1}\bigg\{ \large\frac{(x-3/2)}{\sqrt {17}/4}\bigg\}+c$
$\large\frac{1}{\sqrt {17}}$$\tan ^{-1} \bigg\{ \frac{(x-3/2)}{\sqrt {17}/4}\bigg\}+c$
Hence c is the correct answer.
answered Dec 31, 2013 by meena.p
 
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