i need to calculate this: $I=\int_{0}^{1}\frac{x^2}{(1+x)^3}dx$ with the help of this: $F(y)=\int_{0}^{1}\frac{dx}{1+yx}$, and the clue is to use $F''(y)$. So $F''(y)=\int_{0}^{1}\frac{2y^2}{(1+yx)^3}dx$, now i'd like to say that $F''(1)=2I$, but i must be missing something that would secure the $x^2$ in $I$, any ideas? Maybe it's very simple, i just don't see it right now. edit: i'm not sure if i should define $F(y)$ as it is or as $F(y)=\int_{0}^{y}\frac{dx}{1+yx}$
Liebnitz rule integral $I=\int_{0}^{1}\frac{x^2}{(1+x)^3}dx$
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integration
derivatives
1 Answers
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You've made a mistake. You should have
$$F''(y)=\int_0^1\frac{2x^2}{(1+yx)^3}\ dx$$
which is due to the chain rule:
$$\frac d{dy}\frac1{1+yx}=\frac{-x}{(1+yx)^2}$$
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0My bad, i thought the chain rule was d/dx – 2017-01-16
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0@CodeHoarder No, it doesn't really make sense to take $\frac d{dx}F(y)$ :-) – 2017-01-16
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0Sorry, i'm just pretty confused on this subject. Thanks for the comment it straightened things up for me. – 2017-01-16
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0@CodeHoarder Ok :-) Glad to straighten things up then. – 2017-01-16