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प्रश्न
Integrate the following with respect to x:
`6/(1 + (3x + 2)^2) - 12/sqrt(1 - (3 - 4x)^2`
उत्तर
`int [6/(1 + (3x + 2)^2) - 12/sqrt(1 - (3 - 4x)^2)] "d"x`
= `6int 1/(1 + (3x + 2)^2) "d"x - 12int 1/sqrt(1 - (3 - 4x)^2) "d"x`
= `6/3 tan^-1 (3x + 2) (- 12)/(- 4) sin^-1 (3 - 4x) + "c"`
= `2 tan^-1 (3x + 2) + 3sin^-1 (3 - 4x) + "c"`
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