What is the integral of t^2sqrt(7t-5) ?

The integral of t^2sqrt(7t-5) is 1/343 (2/7 (7t-5)^{7/2}+4(7t-5)^{5/2}+50/3 (7t-5)^{3/2})+C

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integral of t^2sqrt(7t-5)

\int t^{2} 7 t - 5 d t

\frac{1}{343} (\frac{2}{7} (7 t - 5)^{\frac{7}{2}} + 4 (7 t - 5)^{\frac{5}{2}} + \frac{50}{3} (7 t - 5)^{\frac{3}{2}}) + C

Solution steps

\int t^{2} 7 t - 5 d t

Apply u-substitution

: \int \frac{u ( u + 5 ) ^{2}}{343} d u

= \int \frac{u ( u + 5 ) ^{2}}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int u (u + 5)^{2} d u

Expand

u (u + 5)^{2} : u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 u

= \frac{1}{343} \cdot \int u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 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}{343} (\int u^{\frac{5}{2}} d u + \int 10 u^{\frac{3}{2}} d u + \int 25 u d u)

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

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{1}{343} (\frac{2}{7} u^{\frac{7}{2}} + 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}})

Substitute back

u = 7 t - 5

= \frac{1}{343} (\frac{2}{7} (7 t - 5)^{\frac{7}{2}} + 4 (7 t - 5)^{\frac{5}{2}} + \frac{50}{3} (7 t - 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{343} (\frac{2}{7} (7 t - 5)^{\frac{7}{2}} + 4 (7 t - 5)^{\frac{5}{2}} + \frac{50}{3} (7 t - 5)^{\frac{3}{2}}) + C

x \to 0 + lim (ln ∣ x ∣)

4 sec^{2} (x)

\frac{x ^{2}}{7 + 4 x}

x \to \infty lim (\frac{1}{2})

\int sec^{2} (θ) - 1 d θ

What is the integral of t^2sqrt(7t-5) ?
The integral of t^2sqrt(7t-5) is 1/343 (2/7 (7t-5)^{7/2}+4(7t-5)^{5/2}+50/3 (7t-5)^{3/2})+C