Math  /  Calculus

Question(b) 5(x+2)(x2+1)dx\int \frac{5}{(x+2)\left(x^{2}+1\right)} d x.

Studdy Solution

STEP 1

1. The integral is a rational function that can be solved using partial fraction decomposition.
2. The integral involves a linear factor (x+2)(x+2) and an irreducible quadratic factor (x2+1)(x^2+1).

STEP 2

1. Set up the partial fraction decomposition.
2. Solve for the coefficients in the partial fraction decomposition.
3. Integrate each term separately.

STEP 3

Set up the partial fraction decomposition for the integrand 5(x+2)(x2+1)\frac{5}{(x+2)(x^2+1)}. Assume:
5(x+2)(x2+1)=Ax+2+Bx+Cx2+1\frac{5}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+1}
Multiply through by the common denominator (x+2)(x2+1)(x+2)(x^2+1) to clear the fractions:
5=A(x2+1)+(Bx+C)(x+2)5 = A(x^2+1) + (Bx + C)(x+2)

STEP 4

Expand and collect like terms to solve for AA, BB, and CC:
5=Ax2+A+Bx2+2Bx+Cx+2C5 = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C
Combine like terms:
5=(A+B)x2+(2B+C)x+(A+2C)5 = (A + B)x^2 + (2B + C)x + (A + 2C)
Equate the coefficients of the polynomial to those on the right side:
1. A+B=0A + B = 0
2. 2B+C=02B + C = 0
3. A+2C=5A + 2C = 5

Solve this system of equations:
From equation 1: B=AB = -A
Substitute B=AB = -A into equation 2:
2(A)+C=02A+C=0C=2A2(-A) + C = 0 \Rightarrow -2A + C = 0 \Rightarrow C = 2A
Substitute AA and CC into equation 3:
A+2(2A)=5A+4A=55A=5A=1A + 2(2A) = 5 \Rightarrow A + 4A = 5 \Rightarrow 5A = 5 \Rightarrow A = 1
Thus, B=1B = -1 and C=2C = 2.

STEP 5

Substitute the values of AA, BB, and CC back into the partial fractions and integrate each term separately:
5(x+2)(x2+1)dx=1x+2dx+x+2x2+1dx\int \frac{5}{(x+2)(x^2+1)} \, dx = \int \frac{1}{x+2} \, dx + \int \frac{-x + 2}{x^2+1} \, dx
Integrate each term:
1. 1x+2dx=lnx+2+C1\int \frac{1}{x+2} \, dx = \ln|x+2| + C_1
2. xx2+1dx=12lnx2+1+C2\int \frac{-x}{x^2+1} \, dx = -\frac{1}{2} \ln|x^2+1| + C_2 (using substitution u=x2+1u = x^2+1)
3. 2x2+1dx=2tan1(x)+C3\int \frac{2}{x^2+1} \, dx = 2 \tan^{-1}(x) + C_3
Combine the results:
5(x+2)(x2+1)dx=lnx+212lnx2+1+2tan1(x)+C\int \frac{5}{(x+2)(x^2+1)} \, dx = \ln|x+2| - \frac{1}{2} \ln|x^2+1| + 2 \tan^{-1}(x) + C
where C=C1+C2+C3C = C_1 + C_2 + C_3.
The integral evaluates to:
lnx+212lnx2+1+2tan1(x)+C\ln|x+2| - \frac{1}{2} \ln|x^2+1| + 2 \tan^{-1}(x) + C

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