If ∫dx(x+2)x2+1=aln⁡1+x2+btan−1⁡x+15ln⁡|x+2|+C then

∫dx(x+2)x2+1=aln⁡1+x2+btan−1⁡x+15ln⁡|x+2|+C.Differentiating both sides, we get1(x+2)x2+1=2ax1+x2+b1+x2+15(x+2)⇒1(x+2)x2+1=(x+2)(5b+10ax)+1+x251+x2(x+2)⇒ 5=1+x2+5(b+2ax)(x+2)Comparing the like powers of 'x'on both sides, we get  1+10a=0,b+4a=0,10b+1=5⇒ a=−110,b=25