อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Ne[S]zA
Evaluate
$$\int \dfrac{x^2e^{arctan x}}{\sqrt{1+x^2}} dx$$
ช่วยหน่อยครับ  |
\[
\int {\frac{{x^2 e^{\arctan x} }}{{\sqrt {1 + x^2 } }}dx = x} \sqrt {1 + x^2 } e^{\arctan x} - \int {\left( {\frac{{xe^{\arctan x} }}{{\sqrt {1 + x^2 } }} + \sqrt {1 + x^2 } e^{\arctan x} } \right)dx}
\]
\[
= x\sqrt {1 + x^2 } e^{\arctan x} - \left( {\sqrt {1 + x^2 } e^{\arctan x} - \int {\frac{{e^{\arctan x} }}{{\sqrt {1 + x^2 } }}dx} } \right) - \int {\sqrt {1 + x^2 } e^{\arctan x} dx}
\]
\[
= x\sqrt {1 + x^2 } e^{\arctan x} - \sqrt {1 + x^2 } e^{\arctan x} + \int {\left( {\frac{1}{{\sqrt {1 + x^2 } }} - \sqrt {1 + x^2 } } \right)} e^{\arctan x} dx
\]
\[
= x\sqrt {1 + x^2 } e^{\arctan x} - \sqrt {1 + x^2 } e^{\arctan x} - \int {\frac{{x^2 e^{\arctan x} }}{{\sqrt {1 + x^2 } }}dx}
\]
\[
= \frac{1}{2}\left( {x\sqrt {1 + x^2 } - \sqrt {1 + x^2 } } \right)e^{\arctan x} + c = \frac{{\left( {x^3 - x^2 + x - 1} \right)e^{\arctan x} }}{{2\sqrt {1 + x^2 } }} + c
\]