int (x + 3)({x^2} + 6x + 10)^9 , dx equals | vedclass

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(A) Let $I = \int (x + 3)({x^2} + 6x + 10)^9 \, dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int (2x + 6)({x^2} + 6x + 10)^9 \, dx$. Let $u = x^

Vedclass · Accepted Answer

$\frac{1}{20}({x^2} + 6x + 10)^{10} + c$

Vedclass · Answer

(A) Let $I = \int (x + 3)({x^2} + 6x + 10)^9 \, dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int (2x + 6)({x^2} + 6x + 10)^9 \, dx$. Let $u = x^2 + 6x + 10$. Then $du = (2x + 6) \, dx$. Substituting these into the integral: $I = \frac{1}{2} \int u^9 \, du$. $I = \frac{1}{2} \cdot \frac{u^{10}}{10} + c$. $I = \frac{1}{20} u^{10} + c$. Substituting back $u = x^2 + 6x + 10$: $I = \frac{1}{20} ({x^2} + 6x + 10)^{10} + c$.

$\int (x + 3)({x^2} + 6x + 10)^9 \, dx$ equals

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