What is the integral of t/(t^4+7) ?

The integral of t/(t^4+7) is 1/(2sqrt(7))arctan((t^2)/(sqrt(7)))+C

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integral of t/(t^4+7)

\int \frac{t}{t ^{4} + 7} d t

\frac{1}{2 7} arctan (\frac{t ^{2}}{7}) + C

Solution steps

\int \frac{t}{t ^{4} + 7} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 7 )} d u

= \int \frac{1}{2 ( u ^{2} + 7 )} 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}{u ^{2} + 7} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{7 ( v ^{2} + 1 )} 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}{7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{7} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{7} arctan (\frac{t ^{2}}{7})

Simplify

\frac{1}{2} \cdot \frac{1}{7} arctan (\frac{t ^{2}}{7}) : \frac{1}{2 7} arctan (\frac{t ^{2}}{7})

= \frac{1}{2 7} arctan (\frac{t ^{2}}{7})

Add a constant to the solution

= \frac{1}{2 7} arctan (\frac{t ^{2}}{7}) + C

- x cos (x)

d y = (x + 2 y) d x

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

\frac{\partial}{\partial x} (\frac{x}{( x + y ) ^{2}})

\int x e^{x^{5}} d x

What is the integral of t/(t^4+7) ?
The integral of t/(t^4+7) is 1/(2sqrt(7))arctan((t^2)/(sqrt(7)))+C