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Question
`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`
Solution
Let I = `int ((x^2 + 2)/(x^2 + 1))"a"^(x + tan^(-1_x))"d"x`
Put x + tan−1x = t
Differentiating w.r.t. x, we get
`(1 + 1/(1 + x^2)) "d"x` = dt
∴ `((x^2 + 2)/(x^2 + 1)) "d"x` = dt
∴ I = `int "a"^1 "dt"`
= `"a"^1/log "a" + "c"`
∴ I = `("a"^(x + tan^(-1_x)))/log "a" + "c"`
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