Integrate indefinite integral:

Solution:
1. use substitution
=>

=>

=>

=>


2. decompose into partial fractions


where A=1/2, B=1/2, C=-1

3. Substitute partial fractions and continue


4. back-substitute u=e^x


Note: log(x) stands for natural log, and NOT log10(x)
We can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that

so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is

second part
12x
12x=

x^1, 1=n-1
n=2
rn=12
2r=12
r=6

is the second bit
last part
-16
-16x^0=

0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
Answer:
A: 1
B: 17
C: 26
D: 42
E: 57
Step-by-step explanation:
45 is exactly half of 90
that means 45 is 50% of 90.
It will be the second option i believe