int (x+1)(x+3)(x+2)^7 , dx = . . . . . . + | vedclass

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Question

(A) Let $I = \int (x+1)(x+3)(x+2)^7 \, dx$. Substitute $u = x+2$,then $du = dx$. Also,$x+1 = u-1$ and $x+3 = u+1$. Substituting these into the integra

Vedclass · Accepted Answer

$\frac{(x+2)^{10}}{10} - \frac{(x+2)^8}{8}$

Vedclass · Answer

(A) Let $I = \int (x+1)(x+3)(x+2)^7 \, dx$. Substitute $u = x+2$,then $du = dx$. Also,$x+1 = u-1$ and $x+3 = u+1$. Substituting these into the integral: $I = \int (u-1)(u+1)u^7 \, du$ $I = \int (u^2-1)u^7 \, du$ $I = \int (u^9 - u^7) \, du$ Integrating with respect to $u$: $I = \frac{u^{10}}{10} - \frac{u^8}{8} + C$ Substituting back $u = x+2$: $I = \frac{(x+2)^{10}}{10} - \frac{(x+2)^8}{8} + C$. Thus,the correct option is $A$.

$\int (x+1)(x+3)(x+2)^7 \, dx = $ . . . . . . $+ C$.

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