What is the integral of t/((t^2+1)^{19/2)} ?

The integral of t/((t^2+1)^{19/2)} is -1/(17(t^2+1)^{17/2)}+C

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integral of t/((t^2+1)^{19/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{19}{2}}} d t

- \frac{1}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}} + C

Solution steps

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{19}{2}}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{19}{2}}} d u

= \int \frac{1}{2 u ^{\frac{19}{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}{u ^{\frac{19}{2}}} d u

Apply the Power Rule

: - \frac{2}{17 u ^{\frac{17}{2}}}

= \frac{1}{2} (- \frac{2}{17 u ^{\frac{17}{2}}})

Substitute back

u = t^{2} + 1

= \frac{1}{2} (- \frac{2}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}}) : - \frac{1}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}}

= - \frac{1}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}}

Add a constant to the solution

= - \frac{1}{17 ( t ^{2} + 1 ) ^{\frac{17}{2}}} + C

f (x) = x^{3} sin (x)

x \to \infty lim (x b)

\int \frac{3 x ^{3}}{7 - 9 x ^{4}} d x

\int \frac{1}{x 4 x - 1} d x

What is the integral of t/((t^2+1)^{19/2)} ?
The integral of t/((t^2+1)^{19/2)} is -1/(17(t^2+1)^{17/2)}+C