What is the integral of t^3sqrt(1+t^2) ?

The integral of t^3sqrt(1+t^2) is 1/15 (3t^2-2)(t^2+1)^{3/2}+C

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integral of t^3sqrt(1+t^2)

\int t^{3} 1 + t^{2} d t

\frac{1}{15} (3 t^{2} - 2) (t^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int t^{3} 1 + t^{2} d t

Apply u-substitution

: \int \frac{( u - 1 ) u}{2} d u

= \int \frac{( u - 1 ) u}{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 (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= \frac{1}{2} \cdot \int u^{\frac{3}{2}} - u 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

= \frac{1}{2} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + t^{2}

= \frac{1}{2} (\frac{2}{5} (1 + t^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + t^{2})^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (1 + t^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + t^{2})^{\frac{3}{2}}) : \frac{1}{15} (3 t^{2} - 2) (t^{2} + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 t^{2} - 2) (t^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 t^{2} - 2) (t^{2} + 1)^{\frac{3}{2}} + C

\int x 4^{x} d x

y = (cos (x))^{s i n (x)}

\int \frac{x}{2 x + 1} d x

x \to 2 lim ((3 - x) (x + 1))

\frac{\partial}{\partial x} (\frac{5}{x y} - 4 x^{2} - 2 y^{2})

What is the integral of t^3sqrt(1+t^2) ?
The integral of t^3sqrt(1+t^2) is 1/15 (3t^2-2)(t^2+1)^{3/2}+C