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Evaluate the following integrals: 1) integral (2x+3)/(x2-3x+2) dx 2a) integral (x7 + 4x6 - 2x5 + (1)/(x) + e^x + x) dx

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2 parts
a

x88+4x77−x63+ln⁡∣x∣+ex−cos⁡x+C\frac{x^8}{8} + \frac{4x^7}{7} - \frac{x^6}{3} + \ln|x| + e^x - \cos x + C

b

17arctan⁡(x7)+C\frac{1}{7} \arctan\left(\frac{x}{7}\right) + C

Step-by-step solution

4 steps

Here are the solutions to your test questions.

1) Evaluate ∫2x+3x2−3x+2dx\int \frac{2x+3}{x^2-3x+2} dx

Step 1: Factor the denominator and perform partial fraction decomposition.
The denominator x2−3x+2x^2-3x+2 factors as (x−1)(x−2)(x-1)(x-2).
We set up the partial fractions:

2x+3(x−1)(x−2)=Ax−1+Bx−2\frac{2x+3}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}

Multiplying by (x−1)(x−2)(x-1)(x-2) gives:

2x+3=A(x−2)+B(x−1)2x+3 = A(x-2) + B(x-1)

To find AA, set x=1x=1:

2(1)+3=A(1−2)+B(1−1)  ⟹  5=−A  ⟹  A=−52(1)+3 = A(1-2) + B(1-1) \implies 5 = -A \implies A = -5

To find BB, set x=2x=2:

2(2)+3=A(2−2)+B(2−1)  ⟹  7=B  ⟹  B=72(2)+3 = A(2-2) + B(2-1) \implies 7 = B \implies B = 7

So, the integral becomes:

∫(−5x−1+7x−2)dx\int \left( \frac{-5}{x-1} + \frac{7}{x-2} \right) dx

Step 2: Integrate each term.

∫−5x−1dx+∫7x−2dx=−5ln⁡∣x−1∣+7ln⁡∣x−2∣+C\int \frac{-5}{x-1} dx + \int \frac{7}{x-2} dx = -5 \ln|x-1| + 7 \ln|x-2| + C

The final answer is:

−5ln⁡∣x−1∣+7ln⁡∣x−2∣+C\boxed{-5 \ln|x-1| + 7 \ln|x-2| + C}

2) Evaluate the following:

Part a· 2 steps

∫(x7+4x6−2x5+1x+ex+sin⁡x)dx\int (x^7 + 4x^6 - 2x^5 + \frac{1}{x} + e^x + \sin x) dx

  1. Integrate each term using the power rule, logarithm rule, exponential rule, and trigonometric rule.

    Recall the integration rules:
    ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C (for n≠−1n \neq -1)
    ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C
    ∫exdx=ex+C\int e^x dx = e^x + C
    ∫sin⁡xdx=−cos⁡x+C\int \sin x dx = -\cos x + C

  2. Apply the rules to each term.

    ∫x7dx=x7+17+1=x88\int x^7 dx = \frac{x^{7+1}}{7+1} = \frac{x^8}{8} ∫4x6dx=4x6+16+1=4x77\int 4x^6 dx = 4 \frac{x^{6+1}}{6+1} = 4 \frac{x^7}{7} ∫−2x5dx=−2x5+15+1=−2x66=−x63\int -2x^5 dx = -2 \frac{x^{5+1}}{5+1} = -2 \frac{x^6}{6} = -\frac{x^6}{3} ∫1xdx=ln⁡∣x∣\int \frac{1}{x} dx = \ln|x| ∫exdx=ex\int e^x dx = e^x ∫sin⁡xdx=−cos⁡x\int \sin x dx = -\cos x

    Combining these results and adding the constant of integration CC:

    x88+4x77−x63+ln⁡∣x∣+ex−cos⁡x+C\boxed{\frac{x^8}{8} + \frac{4x^7}{7} - \frac{x^6}{3} + \ln|x| + e^x - \cos x + C}

Part b

17arctan⁡(x7)+C\frac{1}{7} \arctan\left(\frac{x}{7}\right) + C

∫1x2+49dx\int \frac{1}{x^2+49} dx

  1. Recognize the form of the integral.

    This integral is of the form ∫1x2+a2dx\int \frac{1}{x^2+a^2} dx.
    Here, a2=49a^2 = 49, so a=7a = 7.

  2. Apply the standard integration formula for ∫1x2+a2dx\int \frac{1}{x^2+a^2} dx.

    The formula is ∫1x2+a2dx=1aarctan⁡(xa)+C\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C.
    Substituting a=7a=7:

    ∫1x2+49dx=17arctan⁡(x7)+C\int \frac{1}{x^2+49} dx = \frac{1}{7} \arctan\left(\frac{x}{7}\right) + C

    The final answer is:

    17arctan⁡(x7)+C\boxed{\frac{1}{7} \arctan\left(\frac{x}{7}\right) + C}
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