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Chapter 1: Problem 2

Evaluate \(\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x\)

Short Answer

Expert verified
Question: Evaluate the integral $\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x.$ Answer: $2\sqrt{3x^2 - 2x + 1} + C.$

Step by step solution

01

Simplify the expression and identify a substitution

A good substitution to try is \(u = 3x^2 - 2x + 1\), which helps in simplifying the square root in the denominator. Now, let's find the derivative of \(u\) with respect to \(x\), so that we can replace the expression in the numerator: $$\frac{d u}{d x} = \frac{d (3 x^{2}-2 x+1)}{d x} = 6x - 2.$$ Now, let's rewrite the integral in terms of \(u\): $$\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x = \int \frac{(6x-2)du}{\sqrt{u}}$$
02

Simplify the integral in terms of u and evaluate it

Now, let's substitute \((6x-2)dx = du\) and integrate with respect to \(u\): $$\int \frac{(6x-2)du}{\sqrt{u}} = \int u^{-\frac{1}{2}} du.$$ With this new integral, we can apply standard integration techniques to find the antiderivative: $$\int u^{-\frac{1}{2}} du = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{u} + C.$$
03

Substitute back in terms of x and simplify

Now that we've found the antiderivative in terms of \(u\), we need to convert it back to terms of \(x\). Recall that \(u = 3x^2 - 2x + 1\). Substituting \(u\) back into our antiderivative, we get the final solution: $$2\sqrt{u} + C = 2\sqrt{3x^2 - 2x + 1} + C.$$ So, the given integral evaluates to: $$\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x = 2\sqrt{3x^2 - 2x + 1} + C.$$

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