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

The integral of 1/(x(y)^2+1) is 1/(y^2)ln|1+y^2x|+C

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integral of 1/(x(y)^2+1)

\int \frac{1}{x ( y ) ^{2} + 1} d x

\frac{1}{y ^{2}} ln 1 + y^{2} x + C

Solution steps

\int \frac{1}{x y ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{y ^{2} u} d u

= \int \frac{1}{y ^{2} u} d u

Take the constant out:

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

= \frac{1}{y ^{2}} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{y ^{2}} ln ∣ u ∣

Substitute back

u = 1 + y^{2} x

= \frac{1}{y ^{2}} ln 1 + y^{2} x

Add a constant to the solution

= \frac{1}{y ^{2}} ln 1 + y^{2} x + C

x \to 3 lim (6 2 - x)

\int_{2}^{5} x^{2} ln (3 x) d x

x \frac{d y}{d x} + y = sec (x y)

\int x^{7} ln (1 + x) d x

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

What is the integral of 1/(x(y)^2+1) ?
The integral of 1/(x(y)^2+1) is 1/(y^2)ln|1+y^2x|+C