What is the integral of t/((t^2+1)^{21/2)} ?

The integral of t/((t^2+1)^{21/2)} is -1/(19(t^2+1)^{19/2)}+C

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integral of t/((t^2+1)^{21/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{21}{2}}} d t

- \frac{1}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}} + C

Solution steps

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{21}{2}}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{21}{2}}} d u

= \int \frac{1}{2 u ^{\frac{21}{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{21}{2}}} d u

Apply the Power Rule

: - \frac{2}{19 u ^{\frac{19}{2}}}

= \frac{1}{2} (- \frac{2}{19 u ^{\frac{19}{2}}})

Substitute back

u = t^{2} + 1

= \frac{1}{2} (- \frac{2}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}}) : - \frac{1}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}}

= - \frac{1}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}}

Add a constant to the solution

= - \frac{1}{19 ( t ^{2} + 1 ) ^{\frac{19}{2}}} + C

y^{^{'}} + 2 y = 0, y (0) = 0

y = 5 x^{2}, y = x^{2} + 2

x \to 0 lim (x e^{- \frac{1}{x ^{2}}})

\frac{d}{d x} (ln (3 x^{3} + 4 x^{2} + x + 4))

\frac{d}{d t} (\frac{e ^{2 t}}{1 - t ^{2}})

What is the integral of t/((t^2+1)^{21/2)} ?
The integral of t/((t^2+1)^{21/2)} is -1/(19(t^2+1)^{19/2)}+C