Evaluate the following definite integral:
$integral subscript 0 superscript 1 fraction numerator 2 straight x plus 3 over denominator 5 straight x squared plus 1 end fraction dx$

Solution

Let I = $integral subscript 0 superscript 1 fraction numerator 2 straight x plus 3 over denominator 5 straight x squared plus 1 end fraction dx$
$equals space 2 integral subscript 0 superscript 1 fraction numerator xdx over denominator 5 straight x squared plus 1 end fraction plus 3 integral subscript 0 superscript 1 fraction numerator 1 over denominator 5 straight x squared plus 1 end fraction dx space equals space 2 over 10 integral subscript 0 superscript 1 fraction numerator 10 xdx over denominator 5 straight x squared plus 1 end fraction plus 3 over 5 integral subscript 0 superscript 1 fraction numerator 1 over denominator straight x squared plus begin display style 1 fifth end style end fraction dx$
$equals space 1 fifth integral subscript 0 superscript 1 fraction numerator 10 straight x over denominator 5 straight x squared plus 1 end fraction dx plus 3 over 5 integral subscript 0 superscript 1 fraction numerator 1 over denominator straight x squared plus open parentheses begin display style fraction numerator 1 over denominator square root of 5 end fraction end style close parentheses squared end fraction dx$
$equals space 1 fifth open square brackets log left parenthesis 5 straight x squared plus 1 right parenthesis close square brackets subscript 0 superscript 1 plus 3 over 5. fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator square root of 5 end fraction end style end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator begin display style fraction numerator 1 over denominator square root of 5 end fraction end style end fraction close parentheses close square brackets subscript 0 superscript 1$
$equals space 1 fifth left parenthesis log space 6 minus log 1 right parenthesis plus fraction numerator 3 over denominator square root of 5 end fraction left square bracket tan to the power of negative 1 end exponent square root of 5 minus tan to the power of negative 1 end exponent 0 right square bracket equals space 1 fifth left parenthesis log space 6 space minus 0 right parenthesis plus fraction numerator 3 over denominator square root of 5 end fraction left square bracket tan to the power of negative 1 end exponent square root of 5 minus 0 right square bracket space equals space 1 fifth log space 6 space plus space fraction numerator 3 over denominator square root of 5 end fraction tan to the power of negative 1 end exponent square root of 5$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate the following definite integral:
$integral subscript 0 superscript 1 fraction numerator 2 straight x plus 3 over denominator 5 straight x squared plus 1 end fraction dx$

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction