What is the integral of 1/(1+100x^2) ?

The integral of 1/(1+100x^2) is 1/10 arctan(10x)+C

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integral of 1/(1+100x^2)

\int \frac{1}{1 + 100 x ^{2}} d x

\frac{1}{10} arctan (10 x) + C

Solution steps

\int \frac{1}{1 + 100 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{10 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{10} arctan (u)

Substitute back

u = 10 x

= \frac{1}{10} arctan (10 x)

Add a constant to the solution

= \frac{1}{10} arctan (10 x) + C

x \to - 6 lim (\frac{1}{( x + 6 ) ^{2}})

\frac{d}{d x} (\frac{\frac{d}{d x} ( 1 )}{x})

y^{^{''}} - 4 y^{^{'}} - 32 y = 0

\frac{d}{d x} (\frac{x}{4 - x ^{2}})

\frac{d y}{d t} = y (3 - y), y (0) = 4

What is the integral of 1/(1+100x^2) ?
The integral of 1/(1+100x^2) is 1/10 arctan(10x)+C