What is the integral of (100)/(25+16x^2) ?

The integral of (100)/(25+16x^2) is 5arctan(4/5 x)+C

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integral of (100)/(25+16x^2)

\int \frac{100}{25 + 16 x ^{2}} d x

5 arctan (\frac{4}{5} x) + C

Solution steps

\int \frac{100}{25 + 16 x ^{2}} d x

Take the constant out:

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

= 100 \cdot \int \frac{1}{25 + 16 x ^{2}} d x

Apply Integral Substitution

= 100 \cdot \int \frac{1}{20 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 100 \cdot \frac{1}{20} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 100 \cdot \frac{1}{20} arctan (u)

Substitute back

u = \frac{4}{5} x

= 100 \cdot \frac{1}{20} arctan (\frac{4}{5} x)

Simplify

100 \cdot \frac{1}{20} arctan (\frac{4}{5} x) : 5 arctan (\frac{4}{5} x)

= 5 arctan (\frac{4}{5} x)

Add a constant to the solution

= 5 arctan (\frac{4}{5} x) + C

What is the integral of (100)/(25+16x^2) ?
The integral of (100)/(25+16x^2) is 5arctan(4/5 x)+C