What is the integral of (1+16x)/(sqrt(25-x^2)) ?

The integral of (1+16x)/(sqrt(25-x^2)) is (1+16x)arcsin(1/5 x)-16(xarcsin(x/5)+sqrt(25-x^2))+C

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integral of (1+16x)/(sqrt(25-x^2))

\int \frac{1 + 16 x}{25 - x ^{2}} d x

(1 + 16 x) arcsin (\frac{1}{5} x) - 16 (x arcsin (\frac{x}{5}) + 25 - x^{2}) + C

Solution steps

\int \frac{1 + 16 x}{25 - x ^{2}} d x

Apply Integration By Parts

: (1 + 16 x) arcsin (\frac{1}{5} x) - \int 16 arcsin (\frac{1}{5} x) d x

= (1 + 16 x) arcsin (\frac{1}{5} x) - \int 16 arcsin (\frac{1}{5} x) d x

\int 16 arcsin (\frac{1}{5} x) d x = 16 (x arcsin (\frac{x}{5}) + 25 - x^{2})

= (1 + 16 x) arcsin (\frac{1}{5} x) - 16 (x arcsin (\frac{x}{5}) + 25 - x^{2})

Add a constant to the solution

= (1 + 16 x) arcsin (\frac{1}{5} x) - 16 (x arcsin (\frac{x}{5}) + 25 - x^{2}) + C

\frac{d}{d x} (\frac{2 x}{3 y})

x \to 7 lim ((x^{2} - 49) (\frac{( x - 7 )}{∣ x - 7 ∣}))

\frac{d}{d x} (6 x + x)

\int \frac{2}{x 9 - x ^{2}} d x

x \to 0 - lim (\frac{1}{0})

What is the integral of (1+16x)/(sqrt(25-x^2)) ?
The integral of (1+16x)/(sqrt(25-x^2)) is (1+16x)arcsin(1/5 x)-16(xarcsin(x/5)+sqrt(25-x^2))+C