1. Intergration help

1. (5x + 3) / (x^(2) + 2x + 5)

2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

i am struggling to intergrate these two. solution method would be appreciated alot.

2. Originally Posted by bobguy
1. (5x + 3) / (x^(2) + 2x + 5)

2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

i am struggling to intergrate these two. solution method would be appreciated alot.
$\displaystyle \int{\frac{5x + 3}{x^2 + 2x + 5}\,dx} = \frac{5}{2}\int{\frac{2x + \frac{6}{5}}{x^2 + 2x + 5}\,dx}$

$\displaystyle = \frac{5}{2}\int{\frac{2x + 2 - \frac{4}{5}}{x^2 + 2x + 5}\,dx}$

$\displaystyle = \frac{5}{2}\int{\frac{2x + 2}{x^2 + 2x + 5}\,dx} - 2\int{\frac{1}{x^2 + 2x + 5}\,dx}$

$\displaystyle = \frac{5}{2}\int{\frac{2x + 2}{x^2 + 2x + 5}\,dx} - 2\int{\frac{1}{(x + 1)^2 + 4}\,dx}$.

Solve the first integral using a $\displaystyle u$ substitution, and solve the second integral using a trigonometric substitution $\displaystyle x + 1 = 2\tan{\theta}$.

3. Originally Posted by bobguy
1. (5x + 3) / (x^(2) + 2x + 5)

2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

i am struggling to intergrate these two. solution method would be appreciated alot.
$\displaystyle \int{\frac{3x + 6}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} = \frac{3}{2}\int{\frac{2x + 4}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}$

$\displaystyle = \frac{3}{2}\int{\frac{2x + 8 - 4}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}$

$\displaystyle = \frac{3}{2}\int{\frac{2x + 8}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} - 6\int{\frac{1}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}$

$\displaystyle = \frac{3}{2}\int{\frac{2x + 8}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} - 6\int{\frac{1}{[(x + 4)^2 - 9]^{\frac{1}{2}}}\,dx}$

Solve the first integral using a $\displaystyle u$ substitution and solve the second integral using the substitution $\displaystyle x + 4 = 3\cosh{t}$.