What is the integral of x^6+1/(sqrt(1-4x^2)) ?

The integral of x^6+1/(sqrt(1-4x^2)) is (x^7)/7+1/2 arcsin(2x)+C

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integral of x^6+1/(sqrt(1-4x^2))

\int x^{6} + \frac{1}{1 - 4 x ^{2}} d x

\frac{x ^{7}}{7} + \frac{1}{2} arcsin (2 x) + C

Solution steps

\int x^{6} + \frac{1}{1 - 4 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 x^{6} d x + \int \frac{1}{1 - 4 x ^{2}} d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int \frac{1}{1 - 4 x ^{2}} d x = \frac{1}{2} arcsin (2 x)

= \frac{x ^{7}}{7} + \frac{1}{2} arcsin (2 x)

Add a constant to the solution

= \frac{x ^{7}}{7} + \frac{1}{2} arcsin (2 x) + C

e^{- 4 t}

(i)^{^{'}}

\frac{d}{d x} (1 + tan^{2} (x))

\frac{d}{d x} (\frac{1}{( x ^{2} + y ^{2} )})

\frac{\partial}{\partial x} (e^{7 x y})

What is the integral of x^6+1/(sqrt(1-4x^2)) ?
The integral of x^6+1/(sqrt(1-4x^2)) is (x^7)/7+1/2 arcsin(2x)+C