What is the integral of 9/(sqrt(1+x^2)) ?

The integral of 9/(sqrt(1+x^2)) is 9ln|sqrt(x^2+1)+x|+C

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integral of 9/(sqrt(1+x^2))

\int \frac{9}{1 + x ^{2}} d x

9 ln x^{2} + 1 + x + C

Solution steps

\int \frac{9}{1 + x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

\int \frac{1}{1 + x ^{2}} d x = ln x^{2} + 1 + x

= 9 ln x^{2} + 1 + x

Add a constant to the solution

= 9 ln x^{2} + 1 + x + C

\frac{d}{d x} (\frac{3 x ^{2}}{ln ( x )})

\frac{d y}{d x} = (x + y)^{2} - 1, y (3) = 4

\int_{0}^{1} 2 (1 + x)^{8} d x

\frac{( x + 2 )}{( x - 2 )}

x \to 3 lim (\frac{2 ^{x} - 8}{3 - x})

What is the integral of 9/(sqrt(1+x^2)) ?
The integral of 9/(sqrt(1+x^2)) is 9ln|sqrt(x^2+1)+x|+C