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

\int_{8}^{\infty} \frac{1}{x ^{2} + x} d x

y^{^{'}} = (y + 4 x)^{2}

x \to 1 + lim (\frac{x ^{2}}{x ^{2} - 1})

\frac{d y}{d x} = 3 e^{2 x} y^{2}

ln (x + 2)

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