Question

Evaluate the following integrals: $$ \int \sqrt{3 x^{2}-6 x+10} d x $$

Short answer

Based on the step-by-step solution provided above, the final answer for the integral is: $$ \int\sqrt{3(x-1)^2+7}dx = \frac{2}{3}(3(x-1)^2+7)^\frac{3}{2} + C $$ Where C is the constant of integration.

Step-by-step solution

Identify the adequate substitution

Let's represent the expression inside the square root as a square of a binomial. Observe that: $$ 3x^2-6x+10 = 3(x^2-2x)+10 = 3[(x-1)^2-1]+10 = 3(x-1)^2+7 $$ Thus, choose the substitution: $$ u=x-1 $$ So, the integral will become: $$ \int \sqrt{3(u+1)^2+7}du $$

Find the differential of the substitution

To find the differential of the substitution, we'll differentiate u with respect to x: $$ u = x-1 \Rightarrow du = dx $$ Now, the original integral can be rewritten in terms of u: $$ \int \sqrt{3(u+1)^2+7}du $$

Evaluate the simplified integral

Now we have a simpler integral to evaluate: $$ \int \sqrt{3(u+1)^2+7}du $$ Our goal is to find an antiderivative of the integrand, which means finding a function F(u) such that the derivative of F(u) is equal to the given integrand. Observe that: $$ F'(u) = \sqrt{3(u+1)^2+7} $$ Let's consider the function: $$ F(u) = \frac{2}{3}(3(u+1)^2+7)^\frac{3}{2} $$ Take its derivative with respect to u: $$ F'(u) = (3(u+1)^2+7)^\frac{3}{2}(2(3)(u+1)) = \sqrt{3(u+1)^2+7} $$ Since \(F'(u)=\sqrt{3(u+1)^2+7}\), we can prove that our F(u) is an antiderivative of the given integrand. So, we found that: $$ \int\sqrt{3(x-1)^2+7}dx = \frac{2}{3}(3(x-1)^2+7)^\frac{3}{2} + C $$ Where C is the constant of integration.