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evaluate the integral: ∫(8x^2+9x+8)/(x^2+1)dx

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nani2006 | Student, Grade 10 | Honors

Posted September 4, 2013 at 10:02 PM via web

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evaluate the integral: ∫(8x^2+9x+8)/(x^2+1)dx

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mvcdc | Student , Undergraduate | (Level 1) Associate Educator

Posted September 4, 2013 at 10:16 PM (Answer #1)

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To evaluate the indefinite integral, we first perform long division on the integrand to get:

`(8x^2 +9x + 8)/(x^2 + 1) = 8 + (9x)/(x^2 + 1)`

Now, we have:

`int [8 +( (9x)/(x^2 + 1))] dx`

The integral of a sum is just the sum of the integrals. Hence, we get:

`int 8 dx + int (9x)/(x^2 + 1) dx`

Constants can be taken out, and the first term simplifies to `8 int dx` . We now have:

`8 int dx + int (9x)/(x^2 + 1) dx`

Then, we substitute `u = x^2 + 1` . This implies that `du = 2xdx` and `xdx = (du)/2` .``

Making the necessary substitutions:

`8 int dx + int 9/(2u) du`

`8 int dx + (9/2) int (1/u) du`

Now, we can easily evaluate the integrals. First term is just `8 int dx = 8x + C_1` . While second term is just `9/2 int (1/u) du = (9/2) log(u) + C_2.`

Over-all, this gives us:

`8x + 9/2 log(u) + C` where C is the over-all integration constant.

Now, we substitute back `u = x^2 + 1`

Therefore, our desired integral is:

`int (8x^2 + 9x + 8)/(x^2 + 1) dx = 8x + 9/2 log(x^2 + 1) +C`

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mvcdc | Student , Undergraduate | (Level 1) Associate Educator

Posted September 4, 2013 at 11:22 PM (Answer #2)

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*Note that log here is the natural logarithm - by log(u) i mean ln(u).

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