What is the integral of (30)/(1+25x^2) ?

The integral of (30)/(1+25x^2) is 6arctan(5x)+C

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integral of (30)/(1+25x^2)

\int \frac{30}{1 + 25 x ^{2}} d x

6 arctan (5 x) + C

Solution steps

\int \frac{30}{1 + 25 x ^{2}} d x

Take the constant out:

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

= 30 \cdot \int \frac{1}{1 + 25 x ^{2}} d x

Apply Integral Substitution

= 30 \cdot \int \frac{1}{5 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 30 \cdot \frac{1}{5} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 30 \cdot \frac{1}{5} arctan (u)

Substitute back

u = 5 x

= 30 \cdot \frac{1}{5} arctan (5 x)

Simplify

30 \cdot \frac{1}{5} arctan (5 x) : 6 arctan (5 x)

= 6 arctan (5 x)

Add a constant to the solution

= 6 arctan (5 x) + C

\int \frac{3 x + 1}{x ^{3} - x ^{2}} d x

\int e^{2 x} e^{e^{x}} d x

\int arctan (\frac{1}{t}) d t

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

e^{\frac{x}{2}}

What is the integral of (30)/(1+25x^2) ?
The integral of (30)/(1+25x^2) is 6arctan(5x)+C