Welcome back Kheny’s — missed you this week.
Here are the solutions to the integration problems:
1. Evaluate ∫x2−3x+22x+3dx
Step 1: Factor the denominator and decompose the integrand into partial fractions.
The denominator x2−3x+2 factors as (x−1)(x−2).
We set up the partial fraction decomposition:
(x−1)(x−2)2x+3=x−1A+x−2B
Multiplying both sides by (x−1)(x−2) gives:
2x+3=A(x−2)+B(x−1)
To find A, set x=1:
2(1)+3=A(1−2)+B(1−1)⟹5=−A⟹A=−5
To find B, set x=2:
2(2)+3=A(2−2)+B(2−1)⟹7=B
So, the integral becomes:
∫(x−1−5+x−27)dx
Step 2: Integrate each term.
∫x−1−5dx=−5ln∣x−1∣
∫x−27dx=7ln∣x−2∣
Step 3: Combine the results and add the constant of integration.
∫x2−3x+22x+3dx=−5ln∣x−1∣+7ln∣x−2∣+C
The final answer is −5ln∣x−1∣+7ln∣x−2∣+C.
2. Evaluate the following:
a) ∫(x7+4x6−2x5+x1+ex+sinx)dx
Step 1: Integrate each term using the power rule, logarithm rule, exponential rule, and trigonometric rule for integration.
∫xndx=n+1xn+1+C
∫x1dx=ln∣x∣+C
∫exdx=ex+C
∫sinxdx=−cosx+C
Step 2: Apply the integration rules to each term.
∫x7dx=7+1x7+1=8x8
∫4x6dx=46+1x6+1=74x7
∫−2x5dx=−25+1x5+1=−26x6=−3x6
∫x1dx=ln∣x∣
∫exdx=ex
∫sinxdx=−cosx
Step 3: Combine all the integrated terms and add the constant of integration.
∫(x7+4x6−2x5+x1+ex+sinx)dx=8x8+74x7−3x6+ln∣x∣+ex−cosx+C
The final answer is 8x8+74x7−3x6+ln∣x∣+ex−cosx+C.
b) ∫x2+491dx
Step 1: Recognize the form of the integral.
This integral matches the standard form ∫x2+a21dx=a1arctan(ax)+C.
Step 2: Identify the value of a.
In this case, a2=49, so a=7.
Step 3: Apply the formula.
∫x2+491dx=71arctan(7x)+C
The final answer is 71arctan(7x)+C.
Drop the next question!