What is the integral of 5/(16x^2+49) ?

The integral of 5/(16x^2+49) is 5/28 arctan(4/7 x)+C

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integral of 5/(16x^2+49)

\int \frac{5}{16 x ^{2} + 49} d x

\frac{5}{28} arctan (\frac{4}{7} x) + C

Solution steps

\int \frac{5}{16 x ^{2} + 49} 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}{16 x ^{2} + 49} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{28 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{28} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 \cdot \frac{1}{28} arctan (u)

Substitute back

u = \frac{4}{7} x

= 5 \cdot \frac{1}{28} arctan (\frac{4}{7} x)

Simplify

5 \cdot \frac{1}{28} arctan (\frac{4}{7} x) : \frac{5}{28} arctan (\frac{4}{7} x)

= \frac{5}{28} arctan (\frac{4}{7} x)

Add a constant to the solution

= \frac{5}{28} arctan (\frac{4}{7} x) + C

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What is the integral of 5/(16x^2+49) ?
The integral of 5/(16x^2+49) is 5/28 arctan(4/7 x)+C