Question
$\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
✓
Answer
Let $\text{I}=\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
Substituting x + 2 = t and dx = dt, we get
$\text{I}=\int\big[2(\text{t}-2)^2+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2(\text{t}^2+4-4\text{t})+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2\text{t}^2+8-8\text{t}+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\Big(2\text{t}^{\frac{5}{2}}+11\text{t}^{-\frac{1}{2}}-8\text{t}^{\frac{3}{2}}\Big)\text{dt}$
$=\frac{4}{7}\text{t}^{\frac{7}{2}}+\frac{22}{3}\text{t}^{\frac{3}{2}}-\frac{16}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
Substituting x + 2 = t and dx = dt, we get
$\text{I}=\int\big[2(\text{t}-2)^2+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2(\text{t}^2+4-4\text{t})+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2\text{t}^2+8-8\text{t}+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\Big(2\text{t}^{\frac{5}{2}}+11\text{t}^{-\frac{1}{2}}-8\text{t}^{\frac{3}{2}}\Big)\text{dt}$
$=\frac{4}{7}\text{t}^{\frac{7}{2}}+\frac{22}{3}\text{t}^{\frac{3}{2}}-\frac{16}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
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