Question

$Evaluate:\int \frac{(3x+5)dx}{\sqrt{x^2+4x+3}}.$

Answer 1

$\int \frac{(3x+5)dx}{\sqrt{(}{x}^2+4x+3)}$

If u=$x^2+4x+3$.

$du=(2x+4)dx$

$3\int \frac{x+\frac{5}{3}}{\sqrt{(}{x}^2+4x+3)}$

=$\frac{3}{2}\int \frac{2x+\frac{10}{3}}{\sqrt{(}{x}^2+4x+3)}$

=$\frac{3}{2}\int \frac{2x+4-4+\frac{10}{3}}{\sqrt{(}{x}^2+4x+3)}dx$

=$\frac{3}{2}\int \frac{2x+4}{\sqrt{(}{x}^2+4x+3)}dx-\frac{3}{2}\int \frac{\frac{2}{3}}{\sqrt{(}{x}^2+4x+3)}$

=$\frac{3}{2}\int \frac{du}{\sqrt{(}u)}-\int \frac{1}{\sqrt{(}u)}$

$I_1$=$\frac{3}{2}\int (u{)}^{\frac{-1}{2}}du$

$I_1$=$\frac{3}{2}\frac{\sqrt{(}u)}{\frac{1}{2}}+C_1$

=$3\sqrt{(}u)+C_1$

Hence, $I_1=3\sqrt{(}{x}^2+4x+3)+C_1$

$I_2=-\int \frac{1}{\sqrt{(}{x}^2+4x+3)}dx$

=$-\int \frac{dx}{\sqrt{(}{x}^2+4x+4-4+3)}$

=$-\int \frac{1}{\sqrt{(}(x+2{)}^2-(1{)}^2}dx$

=$\int \frac{dx}{\sqrt{(}{x}^2-a^2)}$=$\log \mid x+\sqrt{(}{x}^2-a^2)\mid +C_2=\log \mid x+2+\sqrt{(}x+2{)}^2-1)\mid +C_2$

I=$3\sqrt{(}{x}^2+4x+3)-\log \mid x+2+\sqrt{(}{x}^2+4x+3)\mid +C$