Question

lf $\int (2x^3+3x^2-8x+1)\sqrt{2x+6}dx$ $=\frac{\sqrt{2x+6}}{579}(2x+6)f\left(x\right)+C$ then $f\left(x\right)$
is equal to

• A. $x^{3_{-}}6x^2-91x+297$
• B. $7x^3-3x^2-132x+597$
• C. $70x^3-45x^2-396x+897$
• D. $70x^3-45x^2-132x+597$

Answer 1

Answer: (C)

$70x^3-45x^2-396x+897$

Let $I=\int (2x^3+3x^2-8x+1)\sqrt{2x+6}dx$
Substitute $t=\sqrt{2x+6}\Rightarrow dt=\frac{1}{\sqrt{2x+6}}dx$
$I=\int t^2(\frac{1}{4}{(t^2-6)}^3+\frac{3}{4}{(t^2-6)}^2-4(t^2-6)+1)dt$
$=\int (\frac{t^8}{4}-\frac{15t^6}{4}+14t^4-2t^2)dt$
$=\frac{t^9}{36}-\frac{15t^7}{28}+\frac{14t^5}{5}-\frac{2t^3}{3}+c$
$=\frac{2}{315}{(2x+6)}^{\frac{3}{2}}(70x^3-45x^2-396x+897)+c$