I=∫(5x+3)dx√x2+4x+10
=∫5xdx√x2+4x+10+3∫dx√x2+4x+10
=∫52(2x)dx√x2+4x+10+3∫dx√x2+4x+10
=52∫(2x+4−4)dx√x2+4x+10+3∫dx√x2+4x+10
=52∫(2x+4)dx√x2+4x+10−202∫dx1√x2+4x+10+3∫dx√x2+4x+10
=52∫(2x+4)dx√x2+4x+10−10∫dx√x2+4x+10+3∫dx√x2+4x+10
=52∫(2x+4)dx√x2+4x+10−7∫dx√x2+2×2x+4+6
=52∫(2x+4)dx√x2+4x+10−7∫dx√(x+2)2+(√6)2
Let t=x2+4x+10⇒dt=(2x+4)dx
=52∫t−12dt−7log∣∣∣x+2√(x+2)2+6∣∣∣+c since ∫dx√x2+a2=log∣∣x+√x2+a2∣∣+c
=52t−12+1−12+1−7log∣∣∣x+2√(x+2)2+6∣∣∣+c
=52t1212−7log∣∣∣x+2√(x+2)2+6∣∣∣+c
=52×2√t−7log∣∣∣x+2√(x+2)2+6∣∣∣+c
=5√x2+4x+10t−7log∣∣∣x+2√(x+2)2+6∣∣∣+c ........where t=x2+4x+10