Question
Evaluate the integral.$$\int_{-2}^{2} \ln (x+3) d x$$
Step 1
Here, we let $u = \ln(x+3)$ and $dv = dx$. Then, we find $du = \frac{1}{x+3} dx$ and $v = x$. So, we have: $$\int_{-2}^{2} \ln (x+3) d x = \left[ x \ln (x+3) \right]_{-2}^{2} - \int_{-2}^{2} x \cdot \frac{1}{x+3} dx$$ Show more…
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