Question

Integrate x/x^2+2x+3 dx

Answer 1

hmm use the substitution

\[ \alpha \frac{d}{dx}x^2 + 2x + 3 + \beta = x \]
\[2\alpha x + 2\alpha + \beta = x\]
\[\alpha = \frac{1}{2}\]
\[\beta = -\alpha\times 2 = -1\]

Answer 2

\[\alpha \frac{d}{dx}(x^2 + 2x + 3) + \beta = x\]

Answer 3

\[ \frac{\alpha (x^2 + 2x + 3)' + \beta}{x^2 + 2x + 3}\]
\[ \frac{\alpha(x^2 + 3x +3)'}{x^2 + 2x + 3} + \frac{\beta}{x^2 + 2x + 3}\]

for the first one put denominator = t (substitution)

and for second one you will have to make it hmm \((x + 1)^2 + (\sqrt{2})^2\)

Answer 4

you're gonna get

\[\alpha \ln (x^2 +2x + 3) + C_1\]for the first one

Answer 5

and ummm \[\beta \times\frac{1}{\sqrt2}\times \tan^{-1} \frac{(x +1)}{\sqrt{2}}\]

Answer 6

for the second one

Answer 7

is that the final answer when you use the second one?

Answer 8

thanx a lot I completed the square and made the substitution u = x+1
my final answer was 1/2 ln〖(u〗^2+2)-1/√2 〖tan〗^(-1) u/√2+c where u=x+1