Evaluate the following integral using substitution.

Solution

Let I = $integral subscript 0 superscript 1 fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx$
Put $straight e to the power of straight x space equals space straight y comma space space space space space space space space space space space space therefore space space straight e to the power of straight x dx space equals space dy$
When $straight x equals space 0 comma space space space space space straight y space equals space straight e to the power of 0 space equals space 1$
When $straight x space equals space 1 comma space space space space straight y space equals space straight e to the power of 1 space equals space straight e$
$therefore space space space space space space straight I space equals space integral subscript 1 superscript straight e fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space open square brackets tan to the power of negative 1 end exponent space straight y close square brackets subscript 1 superscript straight e space equals space tan to the power of negative 1 end exponent straight e minus tan to the power of negative 1 end exponent 1 space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight e minus 1 over denominator 1 plus straight e end fraction close parentheses$

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Integrals

Evaluate the following integral using substitution.

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