Question

Integrate ∫(3x+1)sqrt (4x^2+12x+5)dx

Answer 1

$I=\int(3x+1)\sqrt{4x^2+12x+5}dx=\int(3x+1)\sqrt{(2x+3)^2-4}dx=$

$=\int\frac{3}{2}(2x+3)\sqrt{(2x+3)^2-4}-\frac{7}{2}\sqrt{(2x+3)^2-4}dx$

$A=\int\sqrt{(2x+3)^2-4}dx=[u=2x+3;du=2dx]=\frac{1}{2}\int\sqrt{u^2-4}du=$

$=[u=2\sec v;du=2\sec v \tan v dv]=\frac{1}{2}\int2\sec v\sqrt{4\sec^2v-4}\tan vdv=$

$=2\int\sec v\tan^2vdv=2\int\sec v(\sec^2v-1)dv=2\int\sec^3 v-\sec vdv=$

$=\sec v\tan v-\ln(\tan v+ \sec v)=[v=arcsec\frac{u}{2}]=u\sqrt{\frac{u^2}{4}-1}+$

$+2\ln\sqrt{\frac{u^2}{4}-1}+\frac{u}{2}=\frac{1}{4}(2x+3)\sqrt{(2x+3)^2-4}-\ln\frac{\sqrt{(2x+3)^2-4}+2x+3}{2}+const$

$B=\int(2x+3)\sqrt{(2x+3)^2-4}=[u=(2x+3)^2-4;du=4(2x+3)dx]=\frac{1}{4}\int\sqrt{u}du=$

$=\frac{2}{12}u^{3/2}=\frac{1}{6}((2x+3)^2-4)\sqrt{(2x+3)^2-4}+const$

$I=\frac{3}{2}B-\frac{7}{2}A=\frac{1}{4}((2x+3)^2-4)\sqrt{(2x+3)^2-4}-\frac{7}{8}(2x+3)\sqrt{(2x+3)^2-4}+$

$+\frac{7}{2}\ln\frac{\sqrt{(2x+3)^2-4}+2x+3}{2}=\sqrt{(2x+3)^2-4}(x^2+3x+\frac{9}{4}-4-\frac{7}{4}x-\frac{21}{8})+$

$+\frac{7}{2}\ln\frac{\sqrt{(2x+3)^2-4}+2x+3}{2}=\sqrt{(2x+3)^2-4}(x^2+\frac{5x}{4}-\frac{11}{8})+$

$+\frac{7}{2}\ln(\sqrt{(2x+3)^2-4}+2x+3)+const$