Evaluate the integrals in Exercises 69–134. The integrals are ...

Question

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.

121. ∫ (1 + x²) / (1 + x³) dx

Accepted Answer

Compute the integral of 5 x2 minus 4 x + 8 divided by x cubed + 8dx. Now, to solve this, we're going to make use of partial fraction decomposition to split this fraction into more fractions. So let's first change our denominator. We have x cubed plus 8. This is a sum of cubes. We know. That A cubed plus B cubed is equals. 2 A + B multiplied by A2 minus AB + B2. So this then factors to give us X. + 2 multiplied by x2 minus 2 x + 4. We can now use this in our partial fraction decomposition. We have 5 x 2 minus 4 x + 8. Divided by X + 2, X2 minus 2 X + 4. We'll set this equals to a divided by X + 2. Plus BX + C divided by X2 minus 2 X + 4. Now, we'd multiply both sides by the denominator with our factor terms to give us 5 X2 minus 4 x + 8 equals A. Multiplied By X2 minus 2 X + 4. Plus b x + c multiplied by x + 2. Now we just need to simplify by factoring and distributing. We end up getting 5 x 2 minus 4 x + 8. Equals A X2 + BX2 minus 2. A X Plus 2 B X. Plus CX. Plus 4 A + 2 C. Now we can group our like terms together. To have 5 X2 minus 4 X + 8 equals A + B. X2 plus 2 B + C minus 2AX. Plus 4 A + 2 C. Now we just need to solve a system to find all of our coefficients. We can do this by equating coefficients. We have X2 on both sides, and there's a 5 in front of the X2 and A + B in front of the X2. 5 equals a plus b. We also have -4. Equals 2 be Plus C minus 2A. And 8 equals 4A plus 2C. Now we just need to solve our system. Let's use substitution. Our first equation we'll solve for beef. To get be Equals 5 minus a. We'll also solve our 3rd equation to get C. C will be equals 24. -28 Now we substitute into our middle equation. We end up getting 2 multiplied by 5 minus A. Plus 4 minus 2A minus 2A. Equals -4. We can then simplify all of this to get -6A equals -18, or A equals 3. This also gives us B. Which is 5 minus 3 or 2. And we can use both of these to find C. C 4 minus 2 multiplied by 3, which is -2. Now we can substitute all of our coefficients. Into our equation. We end up getting our new equation in the integral. We have the integral of a divided by x + 2, which is 3 divided by x + 2. We'll put a d x at the end plus the integral. Of 2 x minus 2 Divided by x2 minus 2 x + 4DX. Now, we know the integral of 3 divided by X + 2. This is 3 natural log absolute value X + 2. Now our other integral is a use substitution. We will let U equals X2 minus 2X +4, which means DU equals 2X minus 2DX. So we can then rewrite our integral. As the integral of d u. Divided by you And we can find that As natural log absolute value of u plus c. So we then add natural log absolute value of u plus c. Now we know what you is. Which is X2 minus 2 X + 4. This is the final answer to our problem, since there's no more to simplify. 3 natural lag. X + 2 + natural log, X2 minus 2 x + 4 + C. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.121. ∫ (1 + x²) / (1 + x³) dx

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Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.
121. ∫ (1 + x²) / (1 + x³) dx