Find integral 1/5-2x^2+4x dx
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Answer:
∫
5−4x−2x
2
dx
Consider 5−4x−2x
2
=−(2x
2
+4x−5)
=−2(x
2
+2x−
2
5
)
=−2(x
2
+2x+1−1−
2
5
)
=−2((x+1)
2
+
2
−2−5
)
=−2((x+1)
2
+
2
−7
)
=2(
2
7
−(x+1)
2
+)
=2((
2
7
)
2
−(x+1)
2
+)
∫
5−4x−2x
2
dx
=∫
2((
2
7
)
2
−(x+1)
2
+)
dx
=
2
1
∫
((
2
7
)
2
−(x+1)
2
+)
dx
We know that ∫
a
2
−x
2
dx
=sin
−1
a
x
+c
Replace x→x→x+1 and a→
2
7
=
2
1
sin
−1
2
7
x+1
+c
=
2
1
sin
−1
7
2
(x+1)+c
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