What is the integral of sqrt(t)(t^6+t-4) ?

The integral of sqrt(t)(t^6+t-4) is 2/15 t^{15/2}+2/5 t^{5/2}-8/3 t^{3/2}+C

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integral of sqrt(t)(t^6+t-4)

\int t (t^{6} + t - 4) d t

\frac{2}{15} t^{\frac{15}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}} + C

Solution steps

\int t (t^{6} + t - 4) d t

Expand

t (t^{6} + t - 4) : t^{\frac{13}{2}} + t^{\frac{3}{2}} - 4 t

= \int t^{\frac{13}{2}} + t^{\frac{3}{2}} - 4 t d t

Apply the Sum Rule:

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

= \int t^{\frac{13}{2}} d t + \int t^{\frac{3}{2}} d t - \int 4 t d t

\int t^{\frac{13}{2}} d t = \frac{2}{15} t^{\frac{15}{2}}

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

\int 4 t d t = \frac{8}{3} t^{\frac{3}{2}}

= \frac{2}{15} t^{\frac{15}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{15} t^{\frac{15}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}} + C

f (x) = 6 + x^{2}

y^{^{''}} + y = sec (x)

x \to 0 lim (\frac{2}{sin ( 7 x )})

n = 1 \sum \infty n^{\frac{1}{n}}

\int 7 x e^{5 x^{2}} d x

What is the integral of sqrt(t)(t^6+t-4) ?
The integral of sqrt(t)(t^6+t-4) is 2/15 t^{15/2}+2/5 t^{5/2}-8/3 t^{3/2}+C