∫2-3x/9x^2+25 DX

∫( 1 on the top and 0 on the bottom of this symbol) x^2+2x-5/x+3 DX

Please guys I really dont understand this. I would appreciate it if you can explain to me step by step on how I would solve these. Thanks

2. Break this up into...

$2\int{dx\over 9x^2+25} -3\int{xdx\over 9x^2+25}$

The first is an arctangent and the second is a basic sub with $u=9x^2+25$

The second integral is a partial fractions, where you should start by dividing.

3. $\int_0^1 \frac{x^2+ 2x-5}{x+3} dx$

Since the denominator has lower degree than the numerator, start by dividing: $\frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}$

$\int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}$
For that last integral, make the substitution u= x+ 3.

4. Originally Posted by HallsofIvy
$\int_0^1 \frac{x^2+ 2x-5}{x+3} dx$

Since the denominator has lower degree than the numerator, start by dividing: $\frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}$

$\int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}$
For that last integral, make the substitution u= x+ 3.
you have a typo in the upper limit in the first integral

5. Originally Posted by HallsofIvy
$\int_0^1 \frac{x^2+ 2x-5}{x+3} dx$

Since the denominator has lower degree than the numerator, start by dividing: $\frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}$

$\int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}$
For that last integral, make the substitution u= x+ 3.
I factor the numerator and obtain a different result.

$\frac{x^2+ 2x-5}{x+3} = \frac{x(x+2) - 5}{x+3} = \frac{x(x+2+1-1) - 5}{x+3} = \frac{ x(x+3) - x - 5 }{x+3}$

$x - \frac{x+5}{x+3} = x - \frac{(x+3)+2}{x+3} = x - 1 - \frac{2}{x+3}$

I cant see where I've made a mistake?

Edit- I subbed in numbers and found that yours has an error somewhere

$\frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}$

Using x=2

$\frac{4+ 4 - 5}{2+ 3}= 2+ 5+ \frac{10}{2+ 3}$

$\frac{3}{5}= 2+ 5+ \frac{10}{5} = 9$ which is not true obviously

Edit Edit (lol) -

The second you is a partial fractions, where you should start by dividing.
I hate dividing polynomials, if you can I would suggest factoring which can be done in this case. Obviously not everybody will share my hatred for dividing polynomials but I treat them like the annoying cousin: when I have to go see them I will, but not a day before!