What is the integral of (10)/(1+4x^2) ?

The integral of (10)/(1+4x^2) is 5arctan(2x)+C

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integral of (10)/(1+4x^2)

\int \frac{10}{1 + 4 x ^{2}} d x

5 arctan (2 x) + C

Solution steps

\int \frac{10}{1 + 4 x ^{2}} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{1 + 4 x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 10 \cdot \frac{1}{2} arctan (u)

Substitute back

u = 2 x

= 10 \cdot \frac{1}{2} arctan (2 x)

Simplify

10 \cdot \frac{1}{2} arctan (2 x) : 5 arctan (2 x)

= 5 arctan (2 x)

Add a constant to the solution

= 5 arctan (2 x) + C

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

\frac{d}{d x} (\frac{5}{( 2 x - 3 ) ^{4}})

(0, 5), (10, 0)

\frac{\partial}{\partial x} (e^{5 x yz - 3})

\int (\frac{1}{y + 4 - 2}) d y

What is the integral of (10)/(1+4x^2) ?
The integral of (10)/(1+4x^2) is 5arctan(2x)+C