Question

Integrate
(A)∫ (3x+1)√4x^2+12x+5 dx
(B)∫x^2+x+5\(x^2+4)(x+1) dx

Answer 1

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

where $u=2x+3$ hence $I=\frac{3}{4}\int u\sqrt{u^2-4} du-\frac{7}{4}\int\sqrt{u^2-4} du$

if $u=\frac{2}{ \ cost},du=\frac{2\ sint}{ \cos^2t}dt$ then $I=3\int\frac{ \sin^3t}{ \cos^4t} dt-\frac{7}{2}\int \frac{ \sin^2t}{\cos^3t} dt$

as H.B.Dwight Tables of integrals , N.Y.,The Mcmillan corp,1961,f.(452.23),f.(452.39) hence

$I=\frac{1-3\cos^2t}{ \cos^3t}-\frac{7}{2}(\frac{ \ sint}{2\ cos^2t}-\frac{1}{2}\ln\tan(\frac{\pi}{4}+\frac{t}{2}))$ where $t=\arccos\frac{2}{2x+3}$

(B). $I=\int(\frac{1}{x+1}+\frac{1}{x^2+4}) dx= \ln$ |$x+1$ |$+\frac{1}{2}\arctan\frac{x}{2}+C$