Question

Evaluate $integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx.$

Solution

Let I = $integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx space equals space integral subscript 0 superscript 2 fraction numerator 5 straight x over denominator straight x squared plus 4 end fraction dx plus integral subscript 0 superscript 2 fraction numerator 1 over denominator straight x squared plus 4 end fraction dx$
$therefore$    $straight I space equals space straight I subscript 1 plus straight I subscript 2$ ...(1)
Put $straight x squared plus 4 space equals space straight y comma space space space therefore space space 2 space straight x space dx space equals space dy space space space rightwards double arrow space space space straight x space dx space equals space 1 half dy$
When x = 0, y = 0 + 4 = 4 When x = 2, y = 4 + 4 = 8
$therefore$    $straight I subscript 1 space equals space 5 over 2 integral subscript 4 superscript 8 1 over straight y dy space equals space 5 over 2 open square brackets log space straight y close square brackets subscript 4 superscript 8 space equals space 5 over 2 left parenthesis log space 8 space minus space log space 4 right parenthesis space equals 5 over 2 log open parentheses 8 over 4 close parentheses space equals space 5 over 2 log space 2$
   $straight I subscript 2 space equals space integral subscript 0 superscript 2 fraction numerator 1 over denominator straight x squared plus 4 end fraction dx space equals space integral subscript 0 superscript 2 fraction numerator 1 over denominator straight x squared plus left parenthesis 2 right parenthesis squared end fraction dx$
$equals space 1 half open square brackets tan to the power of negative 1 end exponent open parentheses straight x over 2 close parentheses close square brackets subscript 0 superscript 2 space space equals space 1 half left square bracket tan to the power of negative 1 end exponent 1 minus tan to the power of negative 1 end exponent 0 right square bracket space equals space 1 half open square brackets straight pi over 4 minus 0 close square brackets space equals space straight pi over 8$
$therefore$ from (1), $straight I space equals 5 over 2 log space 2 space plus space straight pi over 8$

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Integrals

Evaluate $integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx.$

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