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Math  /  Calculus

Question29. ∫(x2−x+1)dxx2+x\int \frac{\left(x^{2}-x+1\right) d x}{x^{2}+x}

Studdy Solution

STEP 1

1. We are given the integral ∫x2−x+1x2+x dx\int \frac{x^2 - x + 1}{x^2 + x} \, dx.
2. We need to find the antiderivative of the given function.

STEP 2

1. Simplify the integrand by performing polynomial long division if necessary.
2. Decompose the resulting expression into partial fractions.
3. Integrate each term separately.
4. Combine the results to find the complete antiderivative.

STEP 3

First, check if polynomial long division is necessary by comparing the degrees of the numerator and denominator. The degree of the numerator x2−x+1x^2 - x + 1 is 2, and the degree of the denominator x2+xx^2 + x is also 2. Therefore, perform polynomial long division.
Divide x2−x+1x^2 - x + 1 by x2+xx^2 + x.

STEP 4

Perform the division:
1. Divide the leading term of the numerator by the leading term of the denominator: x2x2=1\frac{x^2}{x^2} = 1.
2. Multiply the entire divisor by this result: 1⋅(x2+x)=x2+x1 \cdot (x^2 + x) = x^2 + x.
3. Subtract this from the original numerator: $ (x^2 - x + 1) - (x^2 + x) = -2x + 1 \]
The result of the division is 11 with a remainder of −2x+1-2x + 1.

STEP 5

Rewrite the integral using the result from the division:
∫x2−x+1x2+x dx=∫1 dx+∫−2x+1x2+x dx\int \frac{x^2 - x + 1}{x^2 + x} \, dx = \int 1 \, dx + \int \frac{-2x + 1}{x^2 + x} \, dx
Now, decompose −2x+1x2+x\frac{-2x + 1}{x^2 + x} into partial fractions.

STEP 6

Factor the denominator: x2+x=x(x+1)x^2 + x = x(x + 1).
Set up the partial fraction decomposition:
−2x+1x(x+1)=Ax+Bx+1\frac{-2x + 1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1}
Multiply through by the common denominator x(x+1)x(x + 1) to clear the fractions:
−2x+1=A(x+1)+Bx-2x + 1 = A(x + 1) + Bx

STEP 7

Expand and equate coefficients:
−2x+1=Ax+A+Bx-2x + 1 = Ax + A + Bx
Combine like terms:
−2x+1=(A+B)x+A-2x + 1 = (A + B)x + A
Equate coefficients:
1. A+B=−2A + B = -2
2. A=1A = 1
Solve the system of equations:
From equation 2: A=1A = 1.
Substitute A=1A = 1 into equation 1:
1+B=−2  ⟹  B=−31 + B = -2 \implies B = -3
Thus, the partial fractions are:
−2x+1x(x+1)=1x−3x+1\frac{-2x + 1}{x(x + 1)} = \frac{1}{x} - \frac{3}{x + 1}

STEP 8

Integrate each term separately:
∫1 dx=x\int 1 \, dx = x
∫1x dx=ln⁡∣x∣\int \frac{1}{x} \, dx = \ln |x|
∫−3x+1 dx=−3ln⁡∣x+1∣\int \frac{-3}{x + 1} \, dx = -3 \ln |x + 1|

STEP 9

Combine the results:
∫x2−x+1x2+x dx=x+ln⁡∣x∣−3ln⁡∣x+1∣+C\int \frac{x^2 - x + 1}{x^2 + x} \, dx = x + \ln |x| - 3 \ln |x + 1| + C
where CC is the constant of integration.
The antiderivative is:
x+ln⁡∣x∣−3ln⁡∣x+1∣+C\boxed{x + \ln |x| - 3 \ln |x + 1| + C}

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