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

The integral of t/((3t^2+1)^4) is -1/(18(3t^2+1)^3)+C

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

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

- \frac{1}{18 ( 3 t ^{2} + 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u ^{4}} d u

= \int \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{6} (- \frac{1}{3 u ^{3}})

Substitute back

u = 3 t^{2} + 1

= \frac{1}{6} (- \frac{1}{3 ( 3 t ^{2} + 1 ) ^{3}})

Simplify

\frac{1}{6} (- \frac{1}{3 ( 3 t ^{2} + 1 ) ^{3}}) : - \frac{1}{18 ( 3 t ^{2} + 1 ) ^{3}}

= - \frac{1}{18 ( 3 t ^{2} + 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{18 ( 3 t ^{2} + 1 ) ^{3}} + C

y^{^{'}} + y = cos (t), y (0) = 1

y = lo g_{e} (lo g_{e} (x))

h \to 0 lim (\frac{( h + 2 ) ^{3} - 8}{h})

\int \frac{1}{y} d y

\int_{0}^{1} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

What is the integral of t/((3t^2+1)^4) ?
The integral of t/((3t^2+1)^4) is -1/(18(3t^2+1)^3)+C