Question

26x + 3Jox 4

Answer 1

Consider the given integral,

I= ∫ 0 2 ( 6x+3 x 2 +4 )dx

Now, integrate the function,

∫ ( 6x+3 x 2 +4 )dx =3 ∫ ( 2x+1 x 2 +4 )dx =3 ∫ 2x x 2 +4 dx +3 ∫ 1 x 2 +4 dx =3log( x 2 +4 )+ 3 2 tan −1 x 2 =F( x )

By fundamental theorem of calculus, we get

I=F( 2 )−F( 0 ) =[ 3log( 2 2 +4 )+ 3 2 tan −1 ( 2 2 ) ]−[ 3log4+ 3 2 tan −1 0 ] =3log8+ 3 2 tan −1 1−3log4− 3 2 tan −1 0 =3log2+ 3π 8

Thus, the solution of integral is 3log2+ 3π 8 .