Question

Find the following integral : $\int (2x-3cos\ x+e^x) \, dx$

Answer 1

Answer:

$x^2-3sinx+e^x+c$

Step-by-step explanation:

Concept:

$\int[f(x)+g(x)]\:dx=\int{f(x)}\;dx+\int{g(x)}\;dx$

$Now, \\\\\int[2x-3cosx+e^x]\:dx \\\\=\int{2x}\:dx-\int{3cox}\:dx+\int{e^x}\:dx \\\\=2\int{x}\:dx-3\int{cox}\:dx+\int{e^x}\:dx \\\\=2\frac{x^2}{2}-3sinx+e^x+c\\\\=x^2-3sinx+e^x+c$