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Mathematics
Problem
∫
x
2
+
1
x
4
d
x
=
\int \frac{\sqrt{x^2 + 1}}{x^4}\,dx =
∫
x
4
x
2
+
1
d
x
=
(a)
−
1
3
(
x
2
+
1
)
3
/
2
x
3
+
C
-\frac{1}{3}\frac{(x^2 + 1)^{3/2}}{x^3} + C
−
3
1
x
3
(
x
2
+
1
)
3/2
+
C
(b)
x
3
(
x
2
+
1
)
−
1
/
2
+
C
x^3 (x^2 + 1)^{-1/2} + C
x
3
(
x
2
+
1
)
−
1/2
+
C
(c)
x
2
+
1
x
2
+
C
\frac{\sqrt{x^2 + 1}}{x^2} + C
x
2
x
2
+
1
+
C
(d)
−
1
3
(
x
2
+
1
)
3
/
2
x
2
+
C
-\frac{1}{3}\frac{(x^2 + 1)^{3/2}}{x^2} + C
−
3
1
x
2
(
x
2
+
1
)
3/2
+
C
(a)
−
1
3
(
x
2
+
1
)
3
/
2
x
3
+
C
-\frac{1}{3}\frac{(x^2 + 1)^{3/2}}{x^3} + C
−
3
1
x
3
(
x
2
+
1
)
3/2
+
C
(b)
x
3
(
x
2
+
1
)
−
1
/
2
+
C
x^3 (x^2 + 1)^{-1/2} + C
x
3
(
x
2
+
1
)
−
1/2
+
C
(c)
x
2
+
1
x
2
+
C
\frac{\sqrt{x^2 + 1}}{x^2} + C
x
2
x
2
+
1
+
C
(d)
−
1
3
(
x
2
+
1
)
3
/
2
x
2
+
C
-\frac{1}{3}\frac{(x^2 + 1)^{3/2}}{x^2} + C
−
3
1
x
2
(
x
2
+
1
)
3/2
+
C
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