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Integrate `(x^(2))/((a + bx)^(2))`. |
Answer» Correct Answer - `(1)/(b^(3))(a+bx-2a "log"(a+bx)-(a^(2))/(a+bx)+c)` Let `I=(x^(2))/((a+bx)^(2))` Put `a+bx=t rArr "b dx" =dt` `therefore I=int((t-a)/(b))^(2)/t^(2)*(dt)/(b)=(1)/(b^(3))int((t^(2)-2 at + a^(2))/(t^(2)))dt` `=(1)/(b^(3))int(1-(2a)/(t)+(a^(2))/(t^(2)))dt` `(1)/(b^(3))(t-2 "a log t"-(a^(2))/(t))+c` `=(1)/(b^(3))(a+bx-2 "a log"(a+bx)-(a^(2))/(a+bx)+c)` |
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