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प्रश्न
Evaluate the following integral:
`int (x + 1)^2 log x "d"x`
उत्तर
`int (x + 1)^2 log x "d"x`
We use integration by parts method.
Let u = log x
⇒ du = `1/x "d"x`
dv = `(x + 1)^2 "d"x`
So v = `(x + 1)^3/3`
We have `int (x + 1)^2 log x "d"x = (x + 1)^3/3 log x - int (x + 1)^3/3 (1/x) "d"x`
= `(x + 1)^3/3 log x - 1/3 int ((x^3 + 3x^2 + 3x + 1))/x "d"x`
= `(x + 1)^3/3 log x - 1/3 int (x^2 + 3x + 3 + 1/x) "d"x`
= `1/3[(x + 1)^3 log x - x^3/3 - (3x^2)/2 - 3x - log |x|] + "c"`
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