Question:
Evaluate: $\int(2 x-3)^{5}+\sqrt{3 x+2} d x$
Solution:
Let $I=\int(2 x-3)^{5}+\sqrt{3 x+2}$ then,
$I=\int(2 x-3)^{5}+(3 x+2)^{\frac{1}{2}}$
$=\frac{(2 x-3)^{5+1}}{2(5+1)}+\frac{(3 x+2)^{\frac{1}{2}+1}}{3\left(\frac{1}{2}+1\right)}$
$=\frac{(2 x-3)^{6}}{2(6)}+\frac{(3 x+2)^{\frac{3}{2}}}{3\left(\frac{2}{2}\right)}$
$=\frac{(2 x-3)^{6}}{12}+\frac{2(3 x+2)^{\frac{3}{2}}}{9}$
Hence, $I=\frac{(2 x-3)^{6}}{12}+\frac{2(3 x+2)^{\frac{3}{2}}}{9}+C$
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