Question

\(\int\left(x^{2}-2 x+3\right) \ell n x d x\)

Short answer

Question: Find the integral of the function (x^2 - 2x + 3)ln(x) with respect to x. Answer: The integral of the function (x^2 - 2x + 3)ln(x) with respect to x is given by the expression: $$\left(\dfrac{1}{3}x^3 - x^2 + 3x\right)\ell n x - \left(\dfrac{1}{9}x^3 - \dfrac{1}{2}x^2 +3x\right) + C$$ where C is the constant of integration.

Step-by-step solution

Choose u and dv

To choose our u and dv, we will use the following guideline: choose u as the function that simplifies when differentiated, and choose dv as the function that is easy to integrate. In this case, we choose: $$u = \ell n x \implies du = \dfrac{1}{x}dx$$ $$dv = (x^2 - 2x + 3)dx \implies v = \int(x^2 - 2x + 3)dx = \dfrac{1}{3}x^3 - x^2 + 3x$$

Apply Integration by Parts

Now, we will apply the integration by parts formula: $$\int u dv = uv - \int v du$$ Substitute the values we found in Step 1: $$\int (x^2 - 2x + 3)\ell n x dx = \left(\dfrac{1}{3}x^3 - x^2 + 3x\right)\ell n x - \int \left(\dfrac{1}{3}x^3-x^2+3x\right)\dfrac{1}{x} dx$$ Simplify the integral on the right by dividing each term by x: $$\left(\dfrac{1}{3}x^3 - x^2 + 3x\right)\ell n x - \int \left(\dfrac{1}{3}x^2 - x + 3\right) dx$$

Integrate the Remaining Polynomial

Now, we need to integrate the remaining polynomial: $$\int \left(\dfrac{1}{3}x^2 - x + 3\right) dx= \dfrac{1}{9}x^3 - \dfrac{1}{2}x^2 +3x$$

Write the Final Solution

Substitute the integral back into the equation from Step 2, and add the constant of integration C: $$\int (x^2 - 2x + 3)\ell n x dx = \left(\dfrac{1}{3}x^3 - x^2 + 3x\right)\ell n x - \left(\dfrac{1}{9}x^3 - \dfrac{1}{2}x^2 +3x\right) + C$$ This is the final solution for the given integral.