Calculate the integral: integral from 1 to ² x(ln x+1) dx

Calculate the integral: $\int_1^2 x(\ln x+1) \, dx$.

We will use the Newton-Leibniz formula after finding the antiderivative.

Step 1: Expand the integrand

\[\int_1^2 x(\ln x+1) \, dx = \int_1^2 (x\ln x + x) \, dx = \int_1^2 x\ln x \, dx + \int_1^2 x \, dx\]

Step 2: Find the antiderivative of each term

For the second integral:

\[\int x \, dx = \frac{x^2}{2} + C\]

For the first integral: We need to find $\int x\ln x \, dx$.

We use integration by parts: $\int u \, dv = uv - \int v \, du$.

Let $u = \ln x$ and $dv = x \, dx$, so $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.

\[\int x\ln x \, dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2\ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2\ln x}{2} - \frac{x^2}{4} + C\]

Alternative approach - Recognizing a derivative:

Notice that:

\[\frac{d}{dx}\left[\frac{x^2\ln x}{2}\right] = \frac{1}{2}\left[2x\ln x + x^2 \cdot \frac{1}{x}\right] = x\ln x + \frac{x}{2}\]

And:

\[\frac{d}{dx}\left[\frac{x^2}{4}\right] = \frac{x}{2}\]

So:

\[\frac{d}{dx}\left[\frac{x^2\ln x}{2} + \frac{x^2}{4}\right] = x\ln x + \frac{x}{2} + \frac{x}{2} = x\ln x + x = x(\ln x + 1)\]

Therefore, the antiderivative is:

\[F(x) = \frac{x^2\ln x}{2} + \frac{x^2}{4} = \frac{x^2}{4}(2\ln x + 1)\]

Step 3: Apply the Newton-Leibniz formula

\[\int_1^2 x(\ln x+1) \, dx = \left[\frac{x^2\ln x}{2} + \frac{x^2}{4}\right]_1^2\]

Step 4: Evaluate at the upper limit $x = 2$

\[F(2) = \frac{4\ln 2}{2} + \frac{4}{4} = 2\ln 2 + 1\]

Step 5: Evaluate at the lower limit $x = 1$

\[F(1) = \frac{\ln 1}{2} + \frac{1}{4} = 0 + \frac{1}{4} = \frac{1}{4}\]

(since $\ln 1 = 0$)

Step 6: Calculate the definite integral

\[\int_1^2 x(\ln x+1) \, dx = F(2) - F(1) = \left(2\ln 2 + 1\right) - \frac{1}{4} = 2\ln 2 + \frac{4}{4} - \frac{1}{4} = 2\ln 2 + \frac{3}{4}\]

Numerical approximation:

Using $\ln 2 \approx 0.6931$:

\[2\ln 2 + \frac{3}{4} \approx 1.3863 + 0.75 = 2.1363\]

Final answer:

\[\boxed{\int_1^2 x(\ln x+1) \, dx = 2\ln 2 + \frac{3}{4}}\]

Or equivalently:

\[\boxed{\int_1^2 x(\ln x+1) \, dx = \frac{8\ln 2 + 3}{4}}\]