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CBSE
MATHEMATICS · Single correct · CBSE · Mathematics · Integrals · Integrals
Problem
Find
∫
x
e
2
x
d
x
\displaystyle \int x e^{2x}\,dx
∫
x
e
2
x
d
x
. (a)
x
e
2
x
−
1
2
e
2
x
+
C
x e^{2x} - \dfrac{1}{2} e^{2x} + C
x
e
2
x
−
2
1
e
2
x
+
C
(b)
1
2
x
e
2
x
−
1
4
e
2
x
+
C
\dfrac{1}{2} x e^{2x} - \dfrac{1}{4} e^{2x} + C
2
1
x
e
2
x
−
4
1
e
2
x
+
C
(c)
1
2
x
e
2
x
+
1
4
e
2
x
+
C
\dfrac{1}{2} x e^{2x} + \dfrac{1}{4} e^{2x} + C
2
1
x
e
2
x
+
4
1
e
2
x
+
C
(d)
1
2
e
2
x
(
x
−
1
)
+
C
\dfrac{1}{2} e^{2x}(x - 1) + C
2
1
e
2
x
(
x
−
1
)
+
C
Related theory
Integration by Parts
Integration Techniques
Integrals
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