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प्रश्न
Find `int x^2tan^-1x"d"x`
उत्तर
I = `int x^2tan^-1x"d"x`
= `tan^-1x int x^2 "d"x - int 1/(1 + x^2) * x^3/3 "d"x`
= `x^3/3 tan^-1x - 1/3 int (x - x/(1 + x^2))"d"x`
= `x^3/3 tan^-1x - x^2/6 + 1/6 log|1 + x^2| + "C"`
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