Question

Evaluate $\int (4x+3)\sqrt{x^2+10x+27}dx$
(where $C$ is constant of integration)

• A. $\frac{2}{3}{(x^2+10x+27)}^{5/2}-\frac{13}{2}(x+5)\sqrt{x^2+10x+27}-13\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$
• B. $\frac{2}{3}{(x^2+10x+27)}^{3/2}-\frac{13}{2}(x+5)\sqrt{x^2+10x+27}-17\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$
• C. $\frac{4}{3}{(x^2+10x+27)}^{3/2}+\frac{13}{2}(x+5)\sqrt{x^2+10x+27}-17\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$
• D. $\frac{4}{3}{(x^2+10x+27)}^{3/2}-\frac{17}{2}(x+5)\sqrt{x^2+10x+27}-17\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$

Answer 1

The correct option is D $\frac{4}{3}{(x^2+10x+27)}^{3/2}-\frac{17}{2}(x+5)\sqrt{x^2+10x+27}-17\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$

Let $I=\int (4x+3)\sqrt{x^2+10x+27}dx$
$I=2\int (2x+10)\sqrt{x^2+10x+27}dx+\int (3-20)\sqrt{x^2+10x+27}dx$
Put $x^2+10x+27=t$ in first integral, we get

$\Rightarrow I=2\int \sqrt{t}dt-17\int \sqrt{(x+5{)}^2+2}dx$
​​​​$\Rightarrow I=2\cdot \frac{t^{3/2}}{3/2}-17[\left(\frac{x+5}{2}\right)\sqrt{x^2+10x+27}+\frac{2}{2}\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid ]+C$

$\Rightarrow I=\frac{4}{3}{(x^2+10x+27)}^{3/2}-\frac{17}{2}(x+5)\sqrt{x^2+10x+27}-17\ln \mid (x+5)+\sqrt{x^2+10x+27}\mid +C$