What is the integral of (4t^3-3t^2+4t)/(t^2+1) ?

The integral of (4t^3-3t^2+4t)/(t^2+1) is 2t^2-3t+3arctan(t)+2+C

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integral of (4t^3-3t^2+4t)/(t^2+1)

\int \frac{4 t ^{3} - 3 t ^{2} + 4 t}{t ^{2} + 1} d t

2 t^{2} - 3 t + 3 arctan (t) + 2 + C

Solution steps

\int \frac{4 t ^{3} - 3 t ^{2} + 4 t}{t ^{2} + 1} d t

Expand

\frac{4 t ^{3} - 3 t ^{2} + 4 t}{t ^{2} + 1} : \frac{4 t ^{3}}{t ^{2} + 1} - \frac{3 t ^{2}}{t ^{2} + 1} + \frac{4 t}{t ^{2} + 1}

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{4 t ^{3}}{t ^{2} + 1} d t - \int \frac{3 t ^{2}}{t ^{2} + 1} d t + \int \frac{4 t}{t ^{2} + 1} d t

\int \frac{4 t ^{3}}{t ^{2} + 1} d t = 2 (t^{2} + 1 - ln t^{2} + 1)

\int \frac{3 t ^{2}}{t ^{2} + 1} d t = 3 (- arctan (t) + t)

\int \frac{4 t}{t ^{2} + 1} d t = 2 ln t^{2} + 1

= 2 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 2 ln t^{2} + 1

Simplify

2 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 2 ln t^{2} + 1 : 2 t^{2} - 3 t + 3 arctan (t) + 2

= 2 t^{2} - 3 t + 3 arctan (t) + 2

Add a constant to the solution

= 2 t^{2} - 3 t + 3 arctan (t) + 2 + C

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What is the integral of (4t^3-3t^2+4t)/(t^2+1) ?
The integral of (4t^3-3t^2+4t)/(t^2+1) is 2t^2-3t+3arctan(t)+2+C