Question

Evaluate the following ∫√5−2x+x2dx

Answer 1

Let 3x+5=Ad]dx(5+4x−2x2)+B=A(4–4x)+B→A=−343x+5=Ad]dx(5+4x−2x2)+B=A(4–4x)+B→A=−34 and B=8

I=∫(3x+5)5+4x−2x2−−−−−−−−−−√dxI=∫(3x+5)5+4x−2x2dx

=−34(5+4x−2x2)3232+8∫5+4x−2x2−−−−−−−−−−√dx=−34(5+4x−2x2)3232+8∫5+4x−2x2dx

=−12(5+4x−2x2)32+8I2=−12(5+4x−2x2)32+8I2

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

Let x−1=72−−√sintx−1=72sin⁡t Then dx=72−−√costdtdx=72cos⁡tdt

5+4x−2x2=7cos2t5+4x−2x2=7cos2⁡t

I2=∫72√cos2tdt=722√∫(1+cos2t)dt=722√(t+sin2t2)I2=∫72cos2⁡tdt=722∫(1+cos⁡2t)dt=722(t+sin⁡2t2)

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