What is the integral of t/((t^2+1)^{9/2)} ?

The integral of t/((t^2+1)^{9/2)} is -1/(7(t^2+1)^{7/2)}+C

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integral of t/((t^2+1)^{9/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{9}{2}}} d t

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

Solution steps

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{9}{2}}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{9}{2}}} d u

= \int \frac{1}{2 u ^{\frac{9}{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{9}{2}}} d u

Apply the Power Rule

: - \frac{2}{7 u ^{\frac{7}{2}}}

= \frac{1}{2} (- \frac{2}{7 u ^{\frac{7}{2}}})

Substitute back

u = t^{2} + 1

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

Simplify

\frac{1}{2} (- \frac{2}{7 ( t ^{2} + 1 ) ^{\frac{7}{2}}}) : - \frac{1}{7 ( t ^{2} + 1 ) ^{\frac{7}{2}}}

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

Add a constant to the solution

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

y^{^{'}} = 0.5 - t + 2 y

x \to - 7 - lim (\frac{x}{x + 7})

\frac{\frac{1}{2}}{s}

\frac{x}{1 - x}

(y + x + 2) d x + d y = 0

What is the integral of t/((t^2+1)^{9/2)} ?
The integral of t/((t^2+1)^{9/2)} is -1/(7(t^2+1)^{7/2)}+C