Evaluate the following integral using substitution.
$integral subscript 0 superscript 1 fraction numerator straight x over denominator straight x squared plus 1 end fraction dx$

Solution

Let I = $integral subscript 0 superscript 1 fraction numerator straight x over denominator straight x squared plus 1 end fraction dx$
Put $straight x squared plus 1 space equals space straight y semicolon space space space space space space therefore space space 2 straight x space dx space equals space dy semicolon space space space space or space straight x space dx space equals space 1 half dy$
When x = 2, y = 5
When x = 3, y = 10
$therefore space space space straight I space space equals space 1 half integral subscript 5 superscript 10 dy over straight y space equals space 1 half open square brackets log space straight y close square brackets subscript 5 superscript 10 space equals space 1 half left square bracket log space 10 minus space log space 5 right square bracket space equals space 1 half log open parentheses 10 over 5 close parentheses space equals space 1 half log 2$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate the following integral using substitution.
$integral subscript 0 superscript 1 fraction numerator straight x over denominator straight x squared plus 1 end fraction dx$

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction