What is the integral of (7x+2)/(sqrt(1-25x^2)) ?

The integral of (7x+2)/(sqrt(1-25x^2)) is -7/25 sqrt(1-25x^2)+2/5 arcsin(5x)+C

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integral of (7x+2)/(sqrt(1-25x^2))

integral

\frac{7 x + 2}{1 - 25 x ^{2}}

- \frac{7}{25} 1 - 25 x^{2} + \frac{2}{5} arcsin (5 x) + C

Solution steps

\int \frac{7 x + 2}{1 - 25 x ^{2}} d x

Expand

\frac{7 x + 2}{1 - 25 x ^{2}} : \frac{7 x}{1 - 25 x ^{2}} + \frac{2}{1 - 25 x ^{2}}

= \int \frac{7 x}{1 - 25 x ^{2}} + \frac{2}{1 - 25 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{7 x}{1 - 25 x ^{2}} d x + \int \frac{2}{1 - 25 x ^{2}} d x

\int \frac{7 x}{1 - 25 x ^{2}} d x = - \frac{7}{25} 1 - 25 x^{2}

\int \frac{2}{1 - 25 x ^{2}} d x = \frac{2}{5} arcsin (5 x)

= - \frac{7}{25} 1 - 25 x^{2} + \frac{2}{5} arcsin (5 x)

Add a constant to the solution

= - \frac{7}{25} 1 - 25 x^{2} + \frac{2}{5} arcsin (5 x) + C

a^{2} (e^{- a})^{2} x^{4}

\frac{ln ( 4 x )}{x ^{4}}

\frac{1}{( 5 + x ^{2} )}

e^{2} x - 1

(\frac{1}{x - 1})^{\frac{- 1}{2}}

What is the integral of (7x+2)/(sqrt(1-25x^2)) ?
The integral of (7x+2)/(sqrt(1-25x^2)) is -7/25 sqrt(1-25x^2)+2/5 arcsin(5x)+C