Question

ent Scoring st submission for each question part is used for your score. DETAILS PREVIOUS ANSWERS SCALCET8 5.4. 0/1 Points] XY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Find the general indefinite integral. (Use C for the constant of integration.) (7) tv 14 +32x² + at Need Help? Read it Talk to a Tutor Show My Work (Optional) 17,832 30 W

Answer 1

Here we have to find out the given indefinite integrals.

__Part 1)__

Given integral is;

$\int \left ( 7\sqrt {x^3}+4\sqrt[3]{x^2} \right )dx$

So, we started solving it;

(7V73) dx + ſ (1/72]dt

$\int \left (7(x^3)^{\frac{1}{2}} \right )dx+\int (4(x^2)^{\frac{1}{3}})dx$

$=\int \left (7(x)^{\frac{3}{2}} \right )dx+\int (4(x)^{\frac{2}{3}})dx$

=7] dx +4 T (x) dx

$=7 \left [ \dfrac{x^{\frac{3}{2}+1}}{\dfrac{3}{2}+1} \right ] +4 \left [ \dfrac{x^{\frac{2}{3}+1}}{\dfrac{2}{3}+1} \right ]+C$

$=7\left [ \dfrac{x^{\frac{5}{2}}}{\dfrac{5}{2}} \right ]+4\left [ \dfrac{x^{\frac{5}{3}}}{\dfrac{5}{3}} \right ]+C$

$=7\cdot \dfrac{2}{5}\left [ x^{\frac{5}{2}} \right ]+4\cdot \dfrac{3}{5}\left [ x^{\frac{5}{3}} \right ]+C$

$=\dfrac{14}{5}\left [ x^{\frac{5}{2}} \right ]+\dfrac{12}{5}\left [ x^{\frac{5}{3}} \right ]+C$

Here "C" denotes integration constant.

__Part 2)__

Given integral is;

$\int 5v(v^2+2)^2\ dv$

We know;

$\left ( a+b \right )^2=a^2+b^2+2ab$

Hence we get;

$\int 5v\left ( v^4+2^2+2v^2\cdot 2 \right )dv$

$=\int 5v\left ( v^4+4+4v^2 \right )dv$

$=\int \left ( 5v^5+20v+20v^3 \right )dv$

$=\int \left ( 5v^5 \right )dv+\int (20v)\ dv+\int (20v^3)\ dv$

$=5\int v^5\ dv+20\int v\ dv+20\int v^3\ dv$

$=5\left [ \dfrac{v^6}{6} \right ]+20\left [ \dfrac{v^2}{2} \right ]+20\left [ \dfrac{v^4}{4} \right ]+C$

$=\dfrac{5}{6}v^6+10v^2+5v^4+C$

$=5\left [ \dfrac{v^6}{6}+2v^2+v^4 \right ]+C$

Here also "C" represents integration constant.