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∫(x2+2)x2+1ax+tan-1xdx

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प्रश्न

`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`

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उत्तर

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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अध्याय 2.3: Indefinite Integration - Short Answers II

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