What is the integral of (t^3)/(sqrt(t^2+9)) ?

The integral of (t^3)/(sqrt(t^2+9)) is 1/3 (t^2+9)^{3/2}-9sqrt(t^2+9)+C

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integral of (t^3)/(sqrt(t^2+9))

\int \frac{t ^{3}}{t ^{2} + 9} d t

\frac{1}{3} (t^{2} + 9)^{\frac{3}{2}} - 9 t^{2} + 9 + C

Solution steps

\int \frac{t ^{3}}{t ^{2} + 9} d t

Apply u-substitution

: \int u^{2} - 9 d u

= \int u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u - \int 9 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= \frac{u ^{3}}{3} - 9 u

Substitute back

u = t^{2} + 9

= \frac{( t ^{2} + 9 ) ^{3}}{3} - 9 t^{2} + 9

Simplify

\frac{( t ^{2} + 9 ) ^{3}}{3} - 9 t^{2} + 9 : \frac{1}{3} (t^{2} + 9)^{\frac{3}{2}} - 9 t^{2} + 9

= \frac{1}{3} (t^{2} + 9)^{\frac{3}{2}} - 9 t^{2} + 9

Add a constant to the solution

= \frac{1}{3} (t^{2} + 9)^{\frac{3}{2}} - 9 t^{2} + 9 + C

y^{^{'''}} - y^{^{''}} + 4 y^{^{'}} - 4 y = 0

\frac{d}{d x} (\frac{x - 2}{x + 2})

x \to - 2 lim (\frac{∣ x + 2 ∣}{x})

\int e^{- 0.04 x} d x

x \to 500 lim (e^{x})

What is the integral of (t^3)/(sqrt(t^2+9)) ?
The integral of (t^3)/(sqrt(t^2+9)) is 1/3 (t^2+9)^{3/2}-9sqrt(t^2+9)+C