Integrate the function $\frac{8\arctan\left(4y\right)}{1+16y^2}$ from 0 to $\infty $

Step-by-step Solution

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Final answer to the problem

$\frac{\pi ^2}{4}$
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Step-by-step Solution

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  • Integrate by partial fractions
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  • Integrate using tabular integration
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  • Weierstrass Substitution
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  • Integrate using basic integrals
  • Product of Binomials with Common Term
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1

Take out the constant $8$ from the integral

Learn how to solve improper integrals problems step by step online.

$8\int\frac{\arctan\left(4y\right)}{1+16y^2}dy$

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Unlock the first 3 steps of this solution

Learn how to solve improper integrals problems step by step online. Integrate the function (8arctan(4y))/(1+16y^2) from 0 to infinity. Take out the constant 8 from the integral. We can solve the integral \int\frac{\arctan\left(4y\right)}{1+16y^2}dy by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it u), which when substituted makes the integral easier. We see that 4y it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part. Now, in order to rewrite dy in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by deriving the equation above. Isolate dy in the previous equation.

Final answer to the problem

$\frac{\pi ^2}{4}$

Exact Numeric Answer

$2.467401$

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Function Plot

Plotting: $\frac{8\arctan\left(4y\right)}{1+16y^2}$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Improper Integrals

An improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number that is not part of the function's domain, or infinity.

Used Formulas

See formulas (2)

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