Subjects calculus

Integral Substitution Db3F18

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Integral Substitution Db3F18


1. **State the problem:** We want to evaluate the integral $$\int x^2 \sqrt{x+1} \, dx$$ using substitution. 2. **Choose substitution:** Let $$u = x + 1$$. Then, $$du = dx$$ and $$x = u - 1$$. 3. **Rewrite the integral:** Substitute $$x$$ and $$dx$$ in terms of $$u$$: $$\int x^2 \sqrt{x+1} \, dx = \int (u - 1)^2 \sqrt{u} \, du$$ 4. **Expand and simplify:** $$(u - 1)^2 = u^2 - 2u + 1$$ So the integral becomes: $$\int (u^2 - 2u + 1) u^{1/2} \, du = \int (u^{2} u^{1/2} - 2u u^{1/2} + 1 \cdot u^{1/2}) \, du$$ $$= \int (u^{5/2} - 2u^{3/2} + u^{1/2}) \, du$$ 5. **Integrate term-by-term:** Use the power rule $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$: $$\int u^{5/2} du = \frac{u^{7/2}}{7/2} = \frac{2}{7} u^{7/2}$$ $$\int u^{3/2} du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$$ $$\int u^{1/2} du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$$ 6. **Combine results:** $$\int (u^{5/2} - 2u^{3/2} + u^{1/2}) du = \frac{2}{7} u^{7/2} - 2 \cdot \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$$ $$= \frac{2}{7} u^{7/2} - \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$$ 7. **Back-substitute $$u = x + 1$$:** $$\boxed{\frac{2}{7} (x+1)^{7/2} - \frac{4}{5} (x+1)^{5/2} + \frac{2}{3} (x+1)^{3/2} + C}$$ This is the evaluated integral using substitution.