find integral of
x+3/√5-4x+x^2 dx
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Answer:
Step-by-step explanation:
I=∫
x
2
−4x+5
x+3
dx
Puttingx+3=A
dx
d
(x
2
−4x+5)+B
=x+3=A(2x−4)+B
=x+3=2Ax−4A+B
Oncomparingbothsidesweget
2A=1−4A+B=3
A=
2
1
⇒−4×
2
1
+B=3
⇒B−2=3
⇒B=5
I=∫
x
2
−4x+5
2
1
(2x−4)+5
dx
=
2
1
∫
x
2
−4x+5
(2x−4)dx
+5∫
x
2
−4x+5
dx
Put x
2
−4x+5=t
thus,
(2x−4)dx=dt
=
2
1
∫
t
dt
+5∫
x
2
−4x+4−4+5
dx
=
2
1
×2t
2
1
+5∫
(x−2)
2
+1
dx
=
t
+5log
∣
∣
∣
∣
∣
x−2+
(x−2)
2
+1
∣
∣
∣
∣
∣
+c
=
x
2
−4x+5
+5log
∣
∣
∣
x−2+
x
2
−4x+5
∣
∣
∣
+c
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