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74_Quad.pdf
Forms Involving (ax2 + bx + c)
\begin{gather}\int \frac{d x}{\left(a x^{2}+b x+c\right)^{n+1}}=\frac{2 a x+b}{n q\left(a x^{2}+b x+c\right)^{n}}+\frac{2(2 n-1) a}{n q} \int \frac{d x}{\left(a x^{2}+b x+c\right)^{n}} \\ \int \frac{x d x}{\left(a x^{2}+b x+c\right)^{n+1}}=\frac{-(2 c+b x)}{n q\left(a x^{2}+b x+c\right)^{n}}-\frac{(2 n-1) b}{n q} \int \frac{d x}{\left(a x^{2}+b x+c\right)^{n}}\end{gather}其中 $n > 0$ 和 $q = 4ac - b^2$。您可以通过对两侧求导来验证它们。即使将二次因子分解为两个线性因子的乘积，这些公式也是有效的。还可以通过几种方式推出它们。
第一种方法是通过分部积分。将两个积分记为 $I_0(n + 1)$ 和 $I_1(n + 1)$ 并令 $Y = ax^2 + bx + c$。把(1)(2)更简洁地写成$$I_{0}(n+1)=\frac{2 a x+b}{n q Y^{n}}+\frac{2(2 n-1) a I_{0}(n)}{q n} \quad \text { and } \quad I_{1}(n+1)=\frac{-(2 c+b x)}{n q Y^{n}}-\frac{(2 n-1) b I_{0}(n)}{q n}$$首先，对 $I_0(n)$ 部分积分，其中 $u = Y^n$ 和 $dv = dx$ 获得
\begin{align*}
I_0(n)
&= {x\over Y^n} + \int {nx(2ax+b)\,dx\over Y^{n+1}}
= {x\over Y^n} + n\int {2(ax^2+bx+c)-bx-2c\over Y^{n+1}}\,dx
\cr
&= {x\over Y^n} + {2n\over Y^n}
- bnI_1(n+1)
- 2cnI_0(n+1).
\end{align*}
Rearranging, we obtain
\begin{equation}
bnI_1(n+1) + 2cnI_0(n+1) = {x\over Y^n} + (2n-1)I_0(n)
\end{equation}
Second, note that
\begin{equation}
2aI_1(n+1) + bI_0(n) = -1/nY^n
\end{equation}
by using the substitution $t=ax^2=bx+c$.
Equations (3) and (4) can be solved for the two unknowns $I_0(n+1)$ and $I_1(n+1)$.
The second approach is by a trig substitution, provided the quadratic does not factor.
In that case, one can complete the square to obtain
$$
ax^2+bx+c = a\Bigl((x+b/2a)^2 + r^2\Bigr)
~~~{\rm where}~~
r = {\sqrt{4ac-b^2}\over2a}
$$
and $r$ is real.
The substitution $x+b/2 = r\tan t$ converts (1) into
\begin{equation}
\int {\sec^2t\;dt\over (ar^2)^{n+1}\sec^{2n+2}t}
= (4a/q)^{n+1} \int \cos^{2n}t\;dt,
\end{equation}
which can be done by the methods in Section 7.2.
However, to obtain (1), it's necessary to write
\begin{equation}
\int \cos^{2n}t\;dt
= \int \cos^{2n-2}t\;dt - \int \cos^{2n-2}t\;\sin^2t\;dt
\end{equation}
and use integration by parts on the last integral with $u=\sin t$ and $dv=\cos^{2n-2}t\;\cos t\;dt$ from which we have $v = (-\cos{2n-1}t)/(2n-1)$.
After some rearranging and substituting back to eliminate $t$.
In the process, you must remember that (5) is $I_0(n+1)$ so that the middle integral in (6) differs from $I_0(n)$ by factor of $(4a/q)^n$.
The details are left to the reader.
Example
We compute $\int {dx\over(x^2+4)^3}$.
Apply (1) twice, first with $n=2$ and then with $n=1$:
\begin{align}
\int {dx\over(x^2+4)^3}
&= {8x\over32(4x^2+1)^2} + {24\over32}\int {dx\over(4x^2+1)^2}
\cr
&= {x\over4(4x^2+1)^2} + {3\over4}\left(
{8x\over16(4x^2+1)} + {8\over16}\int {dx\over 4x^2+1}\right)
\cr
&= {x\over4(4x^2+1)^2} + {3x\over8(x^2+1)}
+ {3\over8}\,{\arctan 2x\over2} + C,
\end{align}
where the last integral was done by pulling out a factor of 4 from the denominator. |
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