Answer
$3.4 \tan x+\dfrac{1}{ 1.3} \sin x-3.2 e^x +C$
Work Step by Step
We need to simplify: $\int (3.4 \sec^2 x+\dfrac{\cos x}{ 1.3}-3.2 e^x) \ dx $
Separate the integrals.
$\int (3.4 \sec^2 x+\dfrac{\cos x}{ 1.3}-3.2 e^x) \ dx =3.4 \int \sec^2 x dx +\dfrac{1}{ 1.3} \int \cos x dx -3.2 \int e^x \ dx$
or, $=3.4 \tan x+\dfrac{1}{ 1.3} \sin x-3.2 e^x +C$
