What is the integral of (21)/(25+4x^2) ?

The integral of (21)/(25+4x^2) is 21/10 arctan(2/5 x)+C

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integral of (21)/(25+4x^2)

\int \frac{21}{25 + 4 x ^{2}} d x

\frac{21}{10} arctan (\frac{2}{5} x) + C

Solution steps

\int \frac{21}{25 + 4 x ^{2}} d x

Take the constant out:

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

= 21 \cdot \int \frac{1}{25 + 4 x ^{2}} d x

Apply Integral Substitution

= 21 \cdot \int \frac{1}{10 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 21 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 21 \cdot \frac{1}{10} arctan (u)

Substitute back

u = \frac{2}{5} x

= 21 \cdot \frac{1}{10} arctan (\frac{2}{5} x)

Simplify

21 \cdot \frac{1}{10} arctan (\frac{2}{5} x) : \frac{21}{10} arctan (\frac{2}{5} x)

= \frac{21}{10} arctan (\frac{2}{5} x)

Add a constant to the solution

= \frac{21}{10} arctan (\frac{2}{5} x) + C

\int 2 e^{7 - 4 a x} d x

\int (x + y + z) d x

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

\int \frac{5}{u} d u

What is the integral of (21)/(25+4x^2) ?
The integral of (21)/(25+4x^2) is 21/10 arctan(2/5 x)+C