∫dx9x2+6x+5is equal to
tan−13x+12+C
6tan−13x+12+C
cot−13x+12+C
16tan−13x+12+C
∫dx9x2+6x+5=19∫dxx2+23x+59 =19∫dxx2+2⋅13⋅x+132+59−132=19∫dxx2+23x+132+59−19=19∫dxx+132+232=1912/3tan−1x+132/3+C ∵∫dxa2+x2=1atan−1xa =16tan−13x+12+C