Evaluate the following integrals. ∫exe2x+2ex+1dx

Irvin Dukes

Irvin Dukes

Answered

2021-12-20

Evaluate the following integrals.
exe2x+2ex+1dx

Answer & Explanation

kalupunangh

kalupunangh

Expert

2021-12-21Added 29 answers

Step 1
We have to evaluate the integral:
exe2x+2ex+1dx
Rewriting the integral,
exe2x+2ex+1dx=ex(ex)+2ex+1dx
This will be solved by substitution method,
so assuming t=ex
differentiating,
dtdx=dexdx
dtdx=ex
dt=exdx
Step 2
Substituting above values in the integral,
ex(ex)2+2ex+1dx=dtt2+2t+1
=dt(t)2+2(t)(1)+(1)2
=dt(t+1)2
=1t+1+C (since 1(x+a)2=1x+a+C)
Where, C is an arbitrary constant.
Hence, value of integral is 1ex+1+C.
Barbara Meeker

Barbara Meeker

Expert

2021-12-22Added 38 answers

It is required to calculate:
exe2x+2ex+1dx
Substitution u=exdudx=exdx=exdu, we use:
e2x=u2
=1u2+2u+1du
Let us factorize:
=1(u+1)2du
Substitution v=u+1dvdu=1du=dv:
=1v2dv
Integral of a power function:
vndv=vn+1n+1 at n=-2:
=1v
Reverse replacement v=u+1:
=1u+1
Reverse replacement u=ex:
=1ex+1
Problem solved:
exe2x+2ex+1dx
=1ex+1+C
nick1337

nick1337

Expert

2021-12-28Added 573 answers

exe2x+2ex+1dx
Let us put the expression exp (x) under the sign of the differential, ie:
exdx=d(ex),t=ex
Then the original integral can be written as follows:
1t2+2t+1dt
1x2+2x+1dx
Let's
1(x+1)2=1(x+1)2
use the method of decomposition into simplest elements. Let us expand the function into the simplest terms:
1(x+1)2=Ax+1+B(x+1)2=A(x+1)+B(x+1)2
Equate the numerators and take into account that the coefficients at the same powers of x , standing on the left and on the right, must coincide:
1=A(x+1)+B
x: A=0
1: A+B=1
Solving it, we find:
A=0; B=1;
1(x+1)2=0x+1+1(x+1)2
We calculate the table integral:
1(x+1)2dx=1x+1
NSK
Answer:
1x+1+C
To write down the final answer, it remains to substitute exp (x) instead of t.
1ex+1+C

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