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∫
8
x
1
2
x
+
1
d
x
\int 8 x \frac{1}{2 x+1} d x
∫
8
x
2
x
+
1
1
d
x
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Math Problems
Calculus
Find indefinite integrals using the substitution and by parts
Full solution
Q.
∫
8
x
1
2
x
+
1
d
x
\int 8 x \frac{1}{2 x+1} d x
∫
8
x
2
x
+
1
1
d
x
Define u-substitution:
Let's do a u-substitution where
u
=
2
x
+
1
u = 2x + 1
u
=
2
x
+
1
, then
d
u
=
2
d
x
du = 2dx
d
u
=
2
d
x
.
Rewrite in terms of
u
u
u
:
Rewrite the integral in terms of
u
u
u
:
∫
8
x
2
x
+
1
d
x
=
∫
4
u
d
u
\int \frac{8x}{2x+1}\,dx = \int \frac{4}{u} \,du
∫
2
x
+
1
8
x
d
x
=
∫
u
4
d
u
(since
8
x
=
4
(
2
x
)
8x = 4(2x)
8
x
=
4
(
2
x
)
and
2
d
x
=
d
u
2dx = du
2
d
x
=
d
u
).
Integrate
4
/
u
4/u
4/
u
:
Now, integrate
4
u
\frac{4}{u}
u
4
with respect to
u
u
u
:
∫
4
u
d
u
=
4
ln
∣
u
∣
+
C
\int \frac{4}{u} \, du = 4\ln|u| + C
∫
u
4
d
u
=
4
ln
∣
u
∣
+
C
.
Substitute back for u:
Substitute back for u to get the integral in terms of x:
4
ln
∣
2
x
+
1
∣
+
C
4\ln|2x+1| + C
4
ln
∣2
x
+
1∣
+
C
.
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y
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