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Integrate the function $\left(a^{\left(x+1\right)}-2b^{\left(x-1\right)}\right)\left(2b^{\left(x-1\right)}+a^{\left(x+1\right)}\right)$

Step-by-step Solution

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Final answer to the problem

$\frac{a^{\left(2x+2\right)}}{2\ln\left(a\right)}+\frac{-2b^{\left(2x-2\right)}}{\ln\left(b\right)}+C_0$
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Step-by-step Solution

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Find the integral

$\int\left(a^{\left(x+1\right)}-2b^{\left(x-1\right)}\right)\left(2b^{\left(x-1\right)}+a^{\left(x+1\right)}\right)dx$

Learn how to solve integral calculus problems step by step online.

$\int\left(a^{\left(x+1\right)}-2b^{\left(x-1\right)}\right)\left(2b^{\left(x-1\right)}+a^{\left(x+1\right)}\right)dx$

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Learn how to solve integral calculus problems step by step online. Integrate the function (a^(x+1)-2b^(x-1))(2b^(x-1)+a^(x+1)). Find the integral. Rewrite the integrand \left(a^{\left(x+1\right)}-2b^{\left(x-1\right)}\right)\left(2b^{\left(x-1\right)}+a^{\left(x+1\right)}\right) in expanded form. Expand the integral \int\left(a^{\left(2x+2\right)}-4b^{\left(2x-2\right)}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. We can solve the integral \int a^{\left(2x+2\right)}dx 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 2x+2 it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part.

Final answer to the problem

$\frac{a^{\left(2x+2\right)}}{2\ln\left(a\right)}+\frac{-2b^{\left(2x-2\right)}}{\ln\left(b\right)}+C_0$

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Function Plot

Plotting: $\frac{a^{\left(2x+2\right)}}{2\ln\left(a\right)}+\frac{-2b^{\left(2x-2\right)}}{\ln\left(b\right)}+C_0$

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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: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.

Used Formulas

See formulas (4)

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