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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
15
15
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 15
Source: created by kunny
5/15/2005
Calculate the following indefinite integrals. [1]
∫
(
x
2
−
1
)
2
x
4
d
x
\int \frac{(x^2-1)^2}{x^4}dx
∫
x
4
(
x
2
−
1
)
2
d
x
[2]
∫
e
3
x
e
x
+
1
d
x
\int \frac{e^{3x}}{\sqrt{e^x+1}}dx
∫
e
x
+
1
e
3
x
d
x
[3]
∫
sin
2
x
cos
3
x
d
x
\int \sin 2x\cos 3xdx
∫
sin
2
x
cos
3
x
d
x
[4]
∫
x
ln
(
x
+
1
)
d
x
\int x\ln (x+1)dx
∫
x
ln
(
x
+
1
)
d
x
[5]
∫
x
(
x
+
3
)
2
d
x
\int \frac{x}{(x+3)^2}dx
∫
(
x
+
3
)
2
x
d
x
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