Question

integration of √ 1/(2x²+3x+4)

Answer 1

Let 3x+5=Ad]dx(5+4x−2x2)+B=A(4–4x)+B→A=−34 and B=8

I=∫(3x+5)5+4x−2x2−−−−−−−−−−√dx

=−34(5+4x−2x2)3232+8∫5+4x−2x2−−−−−−−−−−√dx

=−12(5+4x−2x2)32+8I2

For I2,5+4x−2x2=7–2(x−1)2=2(72−(x−1)2)

Let x−1=72−−√sint Then dx=72−−√costdt

5+4x−2x2=7cos2t

I2=∫72√cos2tdt=722√∫(1+cos2t)dt=722√(t+sin2t2)

=722√sin−1(27−−√(x−1))+12(x−1)5+4x−2x2−−−−−−−−−−√

I= −12(5+4x−2x2)32+142–√sin−1(27−−√(x−1))+4(x−1)5+4x−2x2−−−−−−−−−−√+C

Answer 2

I hope it helps

thanks ;D