∫ 3 − x 2 ( x 2 + 3 ) 2 d x = − ∫ x 2 − 3 ( x 2 + 3 ) 2 d x = − ∫ ( x 2 + 3 ( x 2 + 3 ) 2 − 6 ( x 2 + 3 ) 2 ) d x = − ∫ ( 1 x 2 + 3 − 6 ( x 2 + 3 ) 2 ) d x = − ∫ 1 x 2 + 3 d x − 6 ∫ 1 ( x 2 + 3 ) 2 d x = − arctan ( x 3 ) 3 + 6 [ x 6 ( x 2 + 3 ) + 1 6 ∫ 1 x 2 + 3 d x ] [ ∵ 1 ( a x 2 + b ) n d x = 2 n − 3 2 b ( n − 1 ) ∫ 1 ( a x 2 + b ) n − 1 d x + x 2 b ( n − 1 ) ( a x 2 + b ) n − 1 ] = − arctan ( x 3 ) 3 + x x 2 + 3 + arctan ( x 3 ) 3 = x x 2 + 3 \begin{aligned}
&\int \frac{3-x^{2}}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\
&=-\int \frac{x^{2}-3}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\
&=-\int\left(\frac{x^{2}+3}{\left(x^{2}+3\right)^{2}}-\frac{6}{\left(x^{2}+3\right)^{2}}\right) \mathrm{d} x \\
&=-\int\left(\frac{1}{x^{2}+3}-\frac{6}{\left(x^{2}+3\right)^{2}}\right) \mathrm{d} x \\
&=-\int \frac{1}{x^{2}+3} \mathrm{~d} x-6 \int \frac{1}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\
&=-\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}+6\left[\frac{x}{6\left(x^{2}+3\right)}+\frac{1}{6} \int \frac{1}{x^{2}+3} \mathrm{~d} x\right] \\
&[\because \frac{1}{\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}}} \mathrm{d} x=\frac{2 \mathrm{n}-3}{2 \mathrm{~b}(\mathrm{n}-1)} \int \frac{1}{\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}-1}} \mathrm{~d} x+\frac{x}{2 \mathrm{~b}(\mathrm{n}-1)\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}-1}} ]\\
&=-\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}+ \frac{x}{x^{2}+3}+\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}
=\frac{x}{x^{2}+3}
\end{aligned} ∫ ( x 2 + 3 ) 2 3 − x 2 d x = − ∫ ( x 2 + 3 ) 2 x 2 − 3 d x = − ∫ ( ( x 2 + 3 ) 2 x 2 + 3 − ( x 2 + 3 ) 2 6 ) d x = − ∫ ( x 2 + 3 1 − ( x 2 + 3 ) 2 6 ) d x = − ∫ x 2 + 3 1 d x − 6 ∫ ( x 2 + 3 ) 2 1 d x = − 3 arctan ( 3 x ) + 6 [ 6 ( x 2 + 3 ) x + 6 1 ∫ x 2 + 3 1 d x ] [ ∵ ( a x 2 + b ) n 1 d x = 2 b ( n − 1 ) 2 n − 3 ∫ ( a x 2 + b ) n − 1 1 d x + 2 b ( n − 1 ) ( a x 2 + b ) n − 1 x ] = − 3 arctan ( 3 x ) + x 2 + 3 x + 3 arctan ( 3 x ) = x 2 + 3 x
#340153 on Dec 2023
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