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Byju's Answer
Standard XII
Mathematics
Factorization
∫ 5 x 2+20 x+...
Question
∫
5
x
2
+
20
x
+
6
x
3
+
2
x
2
+
x
d
x
Open in App
Solution
We
have
,
I
=
∫
5
x
2
+
20
x
+
6
x
3
+
2
x
2
+
x
=
∫
5
x
2
+
20
x
+
6
d
x
x
x
2
+
2
x
+
1
=
∫
5
x
2
+
20
x
+
6
d
x
x
x
+
1
2
Let
5
x
2
+
20
x
+
6
x
x
+
1
2
=
A
x
+
B
x
+
1
+
C
x
+
1
2
⇒
5
x
2
+
20
x
+
6
x
x
+
1
2
=
A
x
+
1
2
+
B
x
x
+
1
+
C
x
x
x
+
1
2
⇒
5
x
2
+
20
x
+
6
=
A
x
2
+
2
x
+
1
+
B
x
2
+
x
+
C
x
⇒
5
x
2
+
20
x
+
6
=
A
+
B
x
2
+
2
A
+
B
+
C
x
+
A
Equating
coefficients
of
like
terms
A
+
B
=
5
.
.
.
.
.
1
2
A
+
B
+
C
=
20
.
.
.
.
.
2
A
=
6
.
.
.
.
.
3
Solving
1
,
2
and
3
,
we
get
A
=
6
B
=
-
1
C
=
9
∴
5
x
2
+
20
x
+
6
x
x
+
1
2
=
6
x
-
1
x
+
1
+
9
x
+
1
2
⇒
I
=
6
∫
d
x
x
-
∫
d
x
x
+
1
+
9
∫
d
x
x
+
1
2
=
6
log
x
-
log
x
+
1
-
9
x
+
1
+
C
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0
Similar questions
Q.
Solve
∫
5
x
2
+
20
x
+
6
x
3
+
2
x
2
+
x
d
x
Q.
Subtract
6
x
3
−
5
x
2
+
4
x
−
2
from the sum of
x
+
2
x
2
−
3
x
3
and
2
−
x
2
+
6
x
−
x
3
Q.
Subtract
4
x
3
+
8
x
2
from
−
2
x
3
+
3
x
2
.
.
Q.
Find the products:
−6x
3
× 5x
2
Q.
6x
3
+ 5x
2
-
21x + 10, 3x
-
2
