What is the integral of (2x+7)/(sqrt(9-4x^2)) ?

The integral of (2x+7)/(sqrt(9-4x^2)) is -1/2 sqrt(9-4x^2)+7/2 arcsin(2/3 x)+C

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integral of (2x+7)/(sqrt(9-4x^2))

\int \frac{2 x + 7}{9 - 4 x ^{2}} d x

- \frac{1}{2} 9 - 4 x^{2} + \frac{7}{2} arcsin (\frac{2}{3} x) + C

Solution steps

\int \frac{2 x + 7}{9 - 4 x ^{2}} d x

Expand

\frac{2 x + 7}{9 - 4 x ^{2}} : \frac{2 x}{9 - 4 x ^{2}} + \frac{7}{9 - 4 x ^{2}}

= \int \frac{2 x}{9 - 4 x ^{2}} + \frac{7}{9 - 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 \frac{2 x}{9 - 4 x ^{2}} d x + \int \frac{7}{9 - 4 x ^{2}} d x

\int \frac{2 x}{9 - 4 x ^{2}} d x = - \frac{1}{2} 9 - 4 x^{2}

\int \frac{7}{9 - 4 x ^{2}} d x = \frac{7}{2} arcsin (\frac{2}{3} x)

= - \frac{1}{2} 9 - 4 x^{2} + \frac{7}{2} arcsin (\frac{2}{3} x)

Add a constant to the solution

= - \frac{1}{2} 9 - 4 x^{2} + \frac{7}{2} arcsin (\frac{2}{3} x) + C

\int_{6}^{8} \frac{60}{( x - 6 ) ^{3}} d x

y = ln (x), at x = 2

\frac{s}{( s ^{2} + 1 ^{2} ) ^{2}}

\int (1 - csc (t) cot (t)) d t

\frac{d}{d x} (a e^{2 x} + b x e^{2 x})

What is the integral of (2x+7)/(sqrt(9-4x^2)) ?
The integral of (2x+7)/(sqrt(9-4x^2)) is -1/2 sqrt(9-4x^2)+7/2 arcsin(2/3 x)+C