Integrate 1
Evaluate
∫ 0 ∞ ( x + x 2 + 1 ) 3 d x
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2 solutions
You should have put
\int_0^\infty
instead of
\int_0^infty
∫
0
i
n
f
t
y
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x
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x
2
+
1
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3
d
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=
∫
0
2
π
(
tan
θ
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tan
2
θ
+
1
)
3
sec
2
θ
d
θ
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⏐
⏐
⏐
⏐
∫
0
∞
(
x
+
x
2
+
1
)
3
d
x
=
∫
0
2
π
(
tan
θ
+
tan
2
θ
+
1
)
3
sec
2
θ
d
θ
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Thanks, I forgot the "\". The braces are unnecessary.
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It gave me an error earlier I think. But now I see that there is no problem
Image is quite unclear....
Let x = tan θ ⇒ x = 0 , ∞ ⇒ θ = 0 , 2 π ⇒ d x = sec 2 θ d θ .
Then, we have:
\begin{aligned} \int_0^\infty \frac{dx}{(x + \sqrt{x^2+1})^3} & = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta}{(\tan \theta + \sqrt{\tan^2 \theta + 1})^3} d \theta \\ & = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta}{(\tan \theta + \sec \theta)^3}d \theta \\ & = \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos^2 \theta}}{(\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} )^3}d \theta \\ & = \int_0^{\frac{\pi}{2}} \frac{\color{#D61F06}{\cos \theta}}{(\color{#3D99F6}{\sin \theta} + 1)^3} \color{#D61F06}{d \theta} \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Let }u = \sin \theta} \quad \Rightarrow \color{#D61F06}{du = \cos \theta \space d \theta} \\ & = \int_\color{#3D99F6}{0}^\color{#3D99F6}{1} \frac{\color{#D61F06}{du}}{(\color{#3D99F6}{u} + 1)^3} \\ & = \left[ -\frac{1}{2(u+1)^2} \right]_0^1 = - \frac{1}{8} + \frac{1}{2} = \boxed{\dfrac{3}{8}} \end{aligned}