What is the integral of 7/(sqrt(3-3x^2)) ?

The integral of 7/(sqrt(3-3x^2)) is 7(x/(sqrt(3-3x^2))-1/(sqrt(3))(-arcsin(x)+x/(sqrt(1-x^2))))+C

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integral of 7/(sqrt(3-3x^2))

\int \frac{7}{3 - 3 x ^{2}} d x

7 (\frac{x}{3 - 3 x ^{2}} - \frac{1}{3} (- arcsin (x) + \frac{x}{1 - x ^{2}})) + C

Solution steps

\int \frac{7}{3 - 3 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int \frac{1}{3 - 3 x ^{2}} d x

Apply Integration By Parts

: \frac{x}{3 - 3 x ^{2}} - \int \frac{3 x ^{2}}{( 3 - 3 x ^{2} ) ^{\frac{3}{2}}} d x

= 7 (\frac{x}{3 - 3 x ^{2}} - \int \frac{3 x ^{2}}{( 3 - 3 x ^{2} ) ^{\frac{3}{2}}} d x)

\int \frac{3 x ^{2}}{( 3 - 3 x ^{2} ) ^{\frac{3}{2}}} d x = \frac{1}{3} (- arcsin (x) + \frac{x}{1 - x ^{2}})

= 7 (\frac{x}{3 - 3 x ^{2}} - \frac{1}{3} (- arcsin (x) + \frac{x}{1 - x ^{2}}))

Add a constant to the solution

= 7 (\frac{x}{3 - 3 x ^{2}} - \frac{1}{3} (- arcsin (x) + \frac{x}{1 - x ^{2}})) + C

\int x (x^{2} - 1)^{3} d x

3 x^{2} - 26 x + 8

f (x) = 2 - 3 x

y^{^{'}} - 3 y = x y^{- 4}

\int tan^{4} (x) sec^{2} (x) d x

What is the integral of 7/(sqrt(3-3x^2)) ?
The integral of 7/(sqrt(3-3x^2)) is 7(x/(sqrt(3-3x^2))-1/(sqrt(3))(-arcsin(x)+x/(sqrt(1-x^2))))+C