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Integrals

Question
CBSEENMA12032401
Wired Faculty App

Evaluate the following definite integral
integral subscript 0 superscript 2 fraction numerator 6 straight x plus 3 over denominator straight x squared plus 4 end fraction dx.

Solution

Let I = integral subscript 0 superscript 2 fraction numerator 6 straight x plus 3 over denominator straight x squared plus 4 end fraction dx space equals space 6 integral subscript 0 superscript 2 fraction numerator straight x over denominator straight x squared plus 4 end fraction dx plus 3 integral subscript 0 superscript 2 fraction numerator 1 over denominator straight x squared plus 4 end fraction dx
       equals 3 integral subscript 0 superscript 2 fraction numerator 2 straight x over denominator straight x squared plus 4 end fraction dx plus 3 integral subscript 0 superscript 2 fraction numerator 1 over denominator straight x squared plus left parenthesis 2 right parenthesis squared end fraction dx space equals space 3 left square bracket log left parenthesis straight x squared plus 4 right parenthesis right square bracket subscript 0 superscript 2 plus 3.1 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 equals 3 left square bracket log space 8 space minus space log space 4 right square bracket space plus 3 over 2 left square bracket tan to the power of negative 1 end exponent 1 space minus space tan to the power of negative 1 end exponent 0 right square bracket space equals space 3 space log open parentheses 8 over 4 close parentheses plus 3 over 2 open parentheses straight pi over 4 minus 0 close parentheses equals space 3 space log space 2 space plus space fraction numerator 3 straight pi over denominator 8 end fraction

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