∫
d
x
4
+
x
2
d
x
\displaystyle \int \frac{dx}{4+x^2} dx
∫
4
+
x
2
d
x
d
x
2
tan
−
1
(
x
2
)
+
C
2\tan^{-1}\left(\dfrac{x}{2}\right)+C
2
tan
−
1
(
2
x
)
+
C
1
2
tan
−
1
(
x
2
)
+
C
\dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right)+C
2
1
tan
−
1
(
2
x
)
+
C
1
2
tan
−
1
(
x
)
+
C
\dfrac{1}{2}\tan^{-1}(x)+C
2
1
tan
−
1
(
x
)
+
C
1
4
tan
−
1
(
x
)
+
C
\dfrac{1}{4}\tan^{-1}(x)+C
4
1
tan
−
1
(
x
)
+
C
Summary
Submit
Approach
The derivative of arctan:
d
d
x
(
tan
−
1
x
)
=
1
1
+
x
2
\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}
d
x
d
(
tan
−
1
x
)
=
1
+
x
2
1
Rewriting our integral:
∫
d
x
4
+
x
2
=
∫
d
x
4
(
1
+
x
2
4
)
\int \frac{dx}{4+x^2} = \int \frac{dx}{4(1+\frac{x^2}{4})}
∫
4
+
x
2
d
x
=
∫
4
(
1
+
4
x
2
)
d
x
=
1
4
∫
d
x
1
+
(
x
2
)
2
d
x
= \frac{1}{4}\int \frac{dx}{1+\left(\frac{x}{2}\right)^2} dx
=
4
1
∫
1
+
(
2
x
)
2
d
x
d
x
Using substitution:
u
=
x
2
d
u
=
d
x
2
2
d
u
=
d
x
u=\frac{x}{2} \hskip{2em} du =\frac{dx}{2} \hskip{2em} 2 du=dx
u
=
2
x
d
u
=
2
d
x
2
d
u
=
d
x
1
4
∫
2
d
u
1
+
u
2
=
1
2
∫
d
u
1
+
u
2
\frac{1}{4}\int \frac{2 du}{1+u^2} =\frac{1}{2}\int \frac{du}{1+u^2}
4
1
∫
1
+
u
2
2
d
u
=
2
1
∫
1
+
u
2
d
u
=
1
2
tan
−
1
u
+
C
= \frac{1}{2}\tan^{-1}u + C
=
2
1
tan
−
1
u
+
C
=
1
2
tan
−
1
(
x
2
)
+
C
= \boxed{\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C}
=
2
1
tan
−
1
(
2
x
)
+
C