Question

Evaluate the following integrals: $$ \int \frac{x+1}{x^{2}+x+3} d x $$

Short answer

Question: Find the exact value of the following integral: \(\int \frac{x+1}{x^{2}+x+3} d x\) Answer: \(\frac{1}{2} \ln |x^2 + x + 3| + C\)

Step-by-step solution

Identify the substitution

We notice that the numerator x+1 is just the derivative of the denominator x^2 + x + 3, or almost a derivative. So we substitute: $$ u=x^{2}+x+3 $$ Now, differentiate u with respect to x: $$ \frac{d u}{d x}=2x+1 $$

Adjust the integral and solve

Now we have du = (2x + 1) dx. Notice that the integral is \(\int \frac{x+1}{u} d x\), so we have \(\frac{1}{2} \int \frac{1}{u} d u.\) Integrate it: $$ \frac{1}{2} \int \frac{1}{u} d u = \frac{1}{2} \ln |u| + C $$

Substitute back and simplify

Now substitute back x in terms of u to get the solution in terms of x (original variable): $$ \frac{1}{2} \ln |x^2 + x + 3| + C $$ So, the result for the integral \(\int \frac{x+1}{x^{2}+x+3} d x\) is $$ \frac{1}{2} \ln |x^2 + x + 3| + C $$