Integrate the function $\frac{e^t}{\sqrt{e^{2t}+9}}$ from 0 to $in^4$

Step-by-step Solution

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Final answer to the problem

$\ln\left|\frac{\sqrt{e^{2in^4}+9}+e^{in^4}}{3}\right|-\ln\left|\frac{\sqrt{e^{0\cdot 2}+9}+e^0}{3}\right|$
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Step-by-step Solution

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  • Integrate by partial fractions
  • Integrate by substitution
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  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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We can solve the integral $\int_{0}^{in^4}\frac{e^t}{\sqrt{e^{2t}+9}}dt$ 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 $e^t$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=e^t$

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$u=e^t$

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Learn how to solve definite integrals problems step by step online. Integrate the function (e^t)/((e^(2t)+9)^(1/2)) from 0 to in^4. We can solve the integral \int_{0}^{in^4}\frac{e^t}{\sqrt{e^{2t}+9}}dt 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 e^t 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 dt 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 dt in the previous equation. Substituting u and dt in the integral and simplify.

Final answer to the problem

$\ln\left|\frac{\sqrt{e^{2in^4}+9}+e^{in^4}}{3}\right|-\ln\left|\frac{\sqrt{e^{0\cdot 2}+9}+e^0}{3}\right|$

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Function Plot

Plotting: $\frac{e^t}{\sqrt{e^{2t}+9}}$

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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: Definite Integrals

Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b

Used Formulas

See formulas (2)

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