What is the integral of t^2sqrt(13t-1) ?

The integral of t^2sqrt(13t-1) is 1/2197 (2/7 (13t-1)^{7/2}+4/5 (13t-1)^{5/2}+2/3 (13t-1)^{3/2})+C

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integral of t^2sqrt(13t-1)

\int t^{2} 13 t - 1 d t

\frac{1}{2197} (\frac{2}{7} (13 t - 1)^{\frac{7}{2}} + \frac{4}{5} (13 t - 1)^{\frac{5}{2}} + \frac{2}{3} (13 t - 1)^{\frac{3}{2}}) + C

Solution steps

\int t^{2} 13 t - 1 d t

Apply u-substitution

: \int \frac{u ( u + 1 ) ^{2}}{2197} d u

= \int \frac{u ( u + 1 ) ^{2}}{2197} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2197} \cdot \int u (u + 1)^{2} d u

Expand

u (u + 1)^{2} : u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u

= \frac{1}{2197} \cdot \int u^{\frac{5}{2}} + 2 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}{2197} (\int u^{\frac{5}{2}} d u + \int 2 u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= \frac{1}{2197} (\frac{2}{7} u^{\frac{7}{2}} + \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 13 t - 1

= \frac{1}{2197} (\frac{2}{7} (13 t - 1)^{\frac{7}{2}} + \frac{4}{5} (13 t - 1)^{\frac{5}{2}} + \frac{2}{3} (13 t - 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{2197} (\frac{2}{7} (13 t - 1)^{\frac{7}{2}} + \frac{4}{5} (13 t - 1)^{\frac{5}{2}} + \frac{2}{3} (13 t - 1)^{\frac{3}{2}}) + C

f (x) = (\frac{4}{x}) + (\frac{1}{5 x ^{3} + 2 x})

\int_{0}^{\frac{π}{2}} sec^{2} (\frac{x}{2}) d x

x y^{^{'}} - 2 y = 0

y^{^{'}} + y = 4 x

\int 2 x + \frac{1}{x ^{2}} + 2 x - 3 d x

What is the integral of t^2sqrt(13t-1) ?
The integral of t^2sqrt(13t-1) is 1/2197 (2/7 (13t-1)^{7/2}+4/5 (13t-1)^{5/2}+2/3 (13t-1)^{3/2})+C