What is the integral of 4x+(4x)/(sqrt(16-x^2)) ?

The integral of 4x+(4x)/(sqrt(16-x^2)) is 2x^2-4sqrt(16-x^2)+C

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integral of 4x+(4x)/(sqrt(16-x^2))

\int 4 x + \frac{4 x}{16 - x ^{2}} d x

2 x^{2} - 4 16 - x^{2} + C

Solution steps

\int 4 x + \frac{4 x}{16 - x ^{2}} 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 4 x d x + \int \frac{4 x}{16 - x ^{2}} d x

\int 4 x d x = 2 x^{2}

\int \frac{4 x}{16 - x ^{2}} d x = - 4 16 - x^{2}

= 2 x^{2} - 4 16 - x^{2}

Add a constant to the solution

= 2 x^{2} - 4 16 - x^{2} + C

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What is the integral of 4x+(4x)/(sqrt(16-x^2)) ?
The integral of 4x+(4x)/(sqrt(16-x^2)) is 2x^2-4sqrt(16-x^2)+C