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

The integral of t/(1+(3t)^4) is 1/18 arctan(9t^2)+C

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

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

\frac{1}{18} arctan (9 t^{2}) + C

Solution steps

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

Simplify

\frac{t}{1 + ( 3 t ) ^{4}} : \frac{t}{1 + 81 t ^{4}}

= \int \frac{t}{1 + 81 t ^{4}} d t

Apply u-substitution

= \int \frac{1}{2 ( 1 + 81 u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{1 + 81 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{9 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{9} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{9} arctan (9 t^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{9} arctan (9 t^{2}) : \frac{1}{18} arctan (9 t^{2})

= \frac{1}{18} arctan (9 t^{2})

Add a constant to the solution

= \frac{1}{18} arctan (9 t^{2}) + C

What is the integral of t/(1+(3t)^4) ?
The integral of t/(1+(3t)^4) is 1/18 arctan(9t^2)+C