Question
Find the integral: $\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x$
✓
Answer
We have
$\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x=\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x$
= $\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C$
= $\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C$
$\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x=\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x$
= $\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C$
= $\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C$
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