∫
8
x
2
−
4
d
x
\displaystyle \int \frac{8}{x^2-4} dx
∫
x
2
−
4
8
d
x
=
4
tan
−
1
(
x
2
)
+
C
4\tan^{-1}\left(\dfrac{x}{2} \right) + C
4
tan
−
1
(
2
x
)
+
C
8
ln
∣
x
2
−
4
∣
+
C
8\ln|x^2-4|+C
8
ln
∣
x
2
−
4∣
+
C
2
ln
∣
x
−
2
x
+
2
∣
+
C
2\ln\left|\dfrac{x-2}{x+2} \right| + C
2
ln
∣
∣
x
+
2
x
−
2
∣
∣
+
C
2
ln
∣
x
+
2
∣
+
2
ln
∣
x
−
2
∣
+
C
2\ln|x+2 |+ 2\ln|x-2| + C
2
ln
∣
x
+
2∣
+
2
ln
∣
x
−
2∣
+
C
Summary
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Approach
Integration by partial fractions:
8
x
2
−
4
=
A
x
+
2
+
B
x
−
2
\frac{8}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2}
x
2
−
4
8
=
x
+
2
A
+
x
−
2
B
A
x
−
2
A
+
B
x
+
2
B
=
8
Ax-2A+Bx+2B=8
A
x
−
2
A
+
B
x
+
2
B
=
8
A
+
B
=
0
−
2
A
+
2
B
=
8
A+B=0 \hskip{2em} -2A+2B=8
A
+
B
=
0
−
2
A
+
2
B
=
8
A
=
−
2
,
B
=
2
A=-2, B=2
A
=
−
2
,
B
=
2
∫
8
x
2
−
4
d
x
=
∫
−
2
x
+
2
+
2
x
−
2
d
x
\int \frac{8}{x^2-4} dx = \int \frac{-2}{x+2} + \frac{2}{x-2} dx
∫
x
2
−
4
8
d
x
=
∫
x
+
2
−
2
+
x
−
2
2
d
x
=
−
2
ln
∣
x
+
2
∣
+
2
ln
∣
x
−
2
∣
+
C
= -2\ln|x+2|+2\ln|x-2|+C
=
−
2
ln
∣
x
+
2∣
+
2
ln
∣
x
−
2∣
+
C
=
2
ln
∣
x
−
2
x
+
2
∣
+
C
= \boxed{2\ln\left|\dfrac{x-2}{x+2} \right| + C }
=
2
ln
∣
∣
x
+
2
x
−
2
∣
∣
+
C