What is the integral of e^x-cos(x)+6x+4 ?

The integral of e^x-cos(x)+6x+4 is e^x-sin(x)+3x^2+4x+C

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integral of e^x-cos(x)+6x+4

\int e^{x} - cos (x) + 6 x + 4 d x

e^{x} - sin (x) + 3 x^{2} + 4 x + C

Solution steps

\int e^{x} - cos (x) + 6 x + 4 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{x} d x - \int cos (x) d x + \int 6 x d x + \int 4 d x

\int e^{x} d x = e^{x}

\int cos (x) d x = sin (x)

\int 6 x d x = 3 x^{2}

\int 4 d x = 4 x

= e^{x} - sin (x) + 3 x^{2} + 4 x

Add a constant to the solution

= e^{x} - sin (x) + 3 x^{2} + 4 x + C

\int - 16 sin (- 4 t) d t

\int \frac{3 x + 2}{2 x} d x

\int \frac{( x - 1 )}{x ^{2} - x - 1} d x

\int \frac{t ^{2}}{t ^{2} + 2} d t

\int \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} d x

What is the integral of e^x-cos(x)+6x+4 ?
The integral of e^x-cos(x)+6x+4 is e^x-sin(x)+3x^2+4x+C