What is the integral of 4/(b^4x^2+m^2) ?

The integral of 4/(b^4x^2+m^2) is 4/(b^2m)arctan((b^2)/m x)+C

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integral of 4/(b^4x^2+m^2)

\int \frac{4}{b ^{4} x ^{2} + m ^{2}} d x

\frac{4}{b ^{2} m} arctan (\frac{b ^{2}}{m} x) + C

Solution steps

\int \frac{4}{b ^{4} x ^{2} + m ^{2}} d x

Take the constant out:

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

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

Apply Integral Substitution

= 4 \cdot \int \frac{1}{b ^{2} m ( u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{b ^{2} m} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 4 \cdot \frac{1}{b ^{2} m} arctan (u)

Substitute back

u = \frac{b ^{2}}{m} x

= 4 \cdot \frac{1}{b ^{2} m} arctan (\frac{b ^{2}}{m} x)

Simplify

4 \cdot \frac{1}{b ^{2} m} arctan (\frac{b ^{2}}{m} x) : \frac{4}{b ^{2} m} arctan (\frac{b ^{2}}{m} x)

= \frac{4}{b ^{2} m} arctan (\frac{b ^{2}}{m} x)

Add a constant to the solution

= \frac{4}{b ^{2} m} arctan (\frac{b ^{2}}{m} x) + C

\int 4 ln (2 x + 1) d x

\frac{d}{d x} ((\frac{cos ( x )}{1 + sin ( x )})^{2})

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

3 x^{2} ln (x), 12 ln (x)

\frac{d}{d x} (sin^{2} (2 x^{3} + 2 x^{2}))

What is the integral of 4/(b^4x^2+m^2) ?
The integral of 4/(b^4x^2+m^2) is 4/(b^2m)arctan((b^2)/m x)+C