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प्रश्न
`int x^3tan^(-1)x "d"x`
उत्तर
Let I = `int x^3*tan^(-1)x*"d"x`
= `int (tan^-1x)x^3 "d"x`
= `(tan^-1x) int x^3 "d"x - int["d"/("d"x) (tan^-1x) int x^3 "d"x] "d"x`
= `(tan^-1x)*(x^4/4) - int 1/(1 + x^2)* x^4/4 "d"x`
= `x^4/4 tan^-1x + 1/4 int ((-x^4))/(1 + x^2) * "d"x`
= `x^4/4 tan^-1x + 1/4 int ((1 - x^4) - 1)/(1 + x) "d"x`
= `x^4/4 tan^-1x + 1/4 int ((1 - x^2)(1 + x^2) - 1)/(1 + x^2) "d"x`
= `x^4/4 tan^-1x + 1/4 int (1 - x^2 - 1/(1 + x^2)) "d"x`
= `x^4/4 tan^-1x + 1/4 (x - x^3/3 - tan^-1x) + "c"`
∴ I = `1/4 tan^-1x (x^4 - 1) - x/12 (x^2 - 3) + "c"`
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