Solve ∫ 3y/1+j^2 wrt y

$$\int \frac { 3 y } { 1 + j ^ { 2 } } d y$$

Evaluate

$$\frac{3y^{2}}{2\left(j^{2}+1\right)}+С$$

Solution Steps

Factor out the constant using \int af\left(y\right)\mathrm{d}y=a\int f\left(y\right)\mathrm{d}y.

$$\frac{3}{1+j^{2}}\int y\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}.

$$\frac{3}{1+j^{2}}\times \frac{y^{2}}{2}$$

Simplify.

$$\frac{3y^{2}}{2\left(1+j^{2}\right)}$$

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{3y^{2}}{2\left(1+j^{2}\right)}+С$$

Differentiate w.r.t. y

$$\frac{3y}{j^{2}+1}$$