What is the integral of (6t-3tx)/(t^2+1) ?

The integral of (6t-3tx)/(t^2+1) is 3ln|1+t^2|-3/2 xln|1+t^2|+C

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integral of (6t-3tx)/(t^2+1)

\int \frac{6 t - 3 t x}{t ^{2} + 1}

3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2} + C

Solution steps

\int \frac{6 t - 3 t x}{t ^{2} + 1} d t

Expand

\frac{6 t - 3 x t}{1 + t ^{2}} : \frac{6 t}{1 + t ^{2}} - \frac{3 x t}{1 + t ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{6 t}{1 + t ^{2}} d t - \int \frac{3 x t}{1 + t ^{2}} d t

\int \frac{6 t}{1 + t ^{2}} d t = 3 ln 1 + t^{2}

\int \frac{3 x t}{1 + t ^{2}} d t = \frac{3}{2} x ln 1 + t^{2}

= 3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2}

Add a constant to the solution

= 3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2} + C

What is the integral of (6t-3tx)/(t^2+1) ?
The integral of (6t-3tx)/(t^2+1) is 3ln|1+t^2|-3/2 xln|1+t^2|+C