What is the integral of 1/(6(x^2+1)) ?

The integral of 1/(6(x^2+1)) is 1/6 arctan(x)+C

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integral of 1/(6(x^2+1))

\int \frac{1}{6 ( x ^{2} + 1 )} d x

\frac{1}{6} arctan (x) + C

Solution steps

\int \frac{1}{6 ( x ^{2} + 1 )} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

\int \frac{1}{x ^{2} + 1} d x = arctan (x)

= \frac{1}{6} arctan (x)

Add a constant to the solution

= \frac{1}{6} arctan (x) + C

f (x) = sin (e^{2 x})

\int \frac{- 2 x + 6}{x ^{2} + 2 x + 3} d x

y^{^{''}} + 8 y^{^{'}} + 15 y = 2 e^{- 5 x}

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 25})

\int e^{t x - x} d x

What is the integral of 1/(6(x^2+1)) ?
The integral of 1/(6(x^2+1)) is 1/6 arctan(x)+C