What is the integral of 5/(x^2+1) ?

The integral of 5/(x^2+1) is 5arctan(x)+C

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integral of 5/(x^2+1)

\int \frac{5}{x ^{2} + 1} d x

5 arctan (x) + C

Solution steps

\int \frac{5}{x ^{2} + 1} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 5 arctan (x)

Add a constant to the solution

= 5 arctan (x) + C

\int sec (\frac{x}{2}) d x

f (x) = 110 x - 16 x^{2}, at 3.438

\frac{x ^{4} - 16 x ^{2} - 16}{( x ^{2} - 4 ) ^{2}}

x y d y - (x^{2} + y^{2}) d x = 0