Integration of x+3/x-1
Answers
Answer:
First step is to distribute the brackets.
⇒
∫
3
1
(
x
+
3
)
(
x
−
1
)
d
x
=
∫
3
1
(
x
2
+
2
x
−
3
)
d
x
Integrate each term using the
power rule for integration
Reminder
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
2
2
∫
(
a
x
n
)
=
a
n
+
1
x
n
+
1
;
n
≠
−
1
2
2
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
=
[
1
3
x
3
+
x
2
−
3
x
]
3
1
Note:
[
F
(
x
)
]
b
a
=
[
F
(
b
)
]
−
[
F
(
a
)
]
⇒
(
9
+
9
−
9
)
−
(
1
3
+
1
−
3
)
=
35
3
Answer link
Explanation:
I’m assuming it’s a general indefinite integral of the form:
y=∫x3x+1dxy=∫x3x+1dx
Now I hope you try it yourself before I start solving.
HINT: Use this
a3+b3=(a+b)(a2−ab+b2)a3+b3=(a+b)(a2−ab+b2)
And these
f′(x)=1x⟺f(x)=ln|x|+C,x≠0f′(x)=1x⟺f(x)=ln|x|+C,x≠0
f′(x)=xn⟺f(x)=xn+1n+1+C,n≠−1f′(x)=xn⟺f(x)=xn+1n+1+C,n≠−1
Hoping you’ve tried, I’m taking a go at it. We have
y=∫x3x+1dxy=∫x3x+1dx
=∫x3+13x+1dx−∫1x+1=∫x3+13x+1dx−∫1x+1