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

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

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

integral

\frac{1}{( 3 t ^{4} + 4 t ^{2} + 1 )}

\frac{3}{2} arctan (3 t) - \frac{1}{2} arctan (t) + C

Solution steps

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

Factor

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

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

Take the partial fraction of

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

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

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

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

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

= \frac{3}{2} arctan (3 t) - \frac{1}{2} arctan (t)

Add a constant to the solution

= \frac{3}{2} arctan (3 t) - \frac{1}{2} arctan (t) + C

\frac{( x - 2 ) ^{3}}{3 x ^{2}}

\frac{t ^{3}}{( 1 + t ^{6} )}

\frac{e ^{3} x}{3}

7 x^{2}

y = 12 (x - 1)^{2} (x + 2)

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