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Question: Solve the given indefinite integral:\[\int {{x^4}{e^{2x}}dx = } \] A. \[\dfrac{{{e^{2x}}}}{4}\left...

Solve the given indefinite integral:∫x4e2xdx=\int {{x^4}{e^{2x}}dx = }
A. e2x4(2x4−4x3+6x2−6x+3)+c\dfrac{{{e^{2x}}}}{4}\left( {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right) + c
B. e2x2(2x4−4x3+6x2−6x+3)+c\dfrac{{{e^{2x}}}}{2}\left( {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right) + c
C. e2x8(2x4+4x3+6x2+6x+3)+c\dfrac{{{e^{2x}}}}{8}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + c
D. e2x4(2x4+4x3+6x2+6x+3)+c\dfrac{{{e^{2x}}}}{4}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + c

Explanation

Solution

Hint: To solve the given question above we will first put 2x=t{\text{2x}} = {\text{t}} and convert whole integrals in terms of t. Now use the integration by parts method to integrate the given function and at last replace t with its original value.

Complete step-by-step answer:
In the question given, we have to find the value of ∫e2x.x4dx\int {{e^{2x}}.{x^4}dx} . Let its value be I. Thus, we have:
I=∫x4e2xdxI = \int {{x^4}{e^{2x}}dx}
Now, we will put 2x=t2x = t .On differentiating both sides we will get:
2dx=dt2dx = dt
⇒dx=dt2\Rightarrow dx = \dfrac{{dt}}{2}
Thus, we will get:
I=∫(t2)4et(dt2)\I=(125)∫et.t4dt\begin{array}{l}I = \int {{{\left( {\dfrac{t}{2}} \right)}^4}{e^t}\left( {\dfrac{{dt}}{2}} \right)} \\\I = \left( {\dfrac{1}{{{2^5}}}} \right)\int {{e^t}.{t^4}dt} \end{array}

Now, we will calculate the value of ∫et.t\int {{e^t}.t} with the help of by parts. According to by parts theorem of integration, we have:
∫uv=u∫v−∫u∫v\int {uv = } u\int {v - \int {u\int v } }
In our case u=t  and  v−etu = t\,\,and\,\,v - {e^t} . Thus, we will get:
∫et.t dt= t∫et dt−∫dtdt∫et (dt)2∫ett dt= tet−∫et (dt)\begin{array}{l}\int {{e^t}.t\,dt = } \,t\int {{e^t}\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {{e^t}\,\left( {dt} \right)} }^2}} } \\\\\int {{e^t}t\,dt = } \,t{e^t} - \int {{e^t}\,\left( {dt} \right)} \end{array}
∫ett dt= tet−et+c1\int {{e^t}t\,dt = } \,t{e^t} - {e^t} + {c_1} . . . . . . . . . . . . . (i)

Now we will calculate the value of ∫ettt dt\int {{e^t}{t^t}\,dt} . In this function, we will again use by-parts. In this case, we will take u=t  and  v=ettu = t\,\,and\,\,v = {e^t}t . Thus, we will get:
∫ett2 dt= t∫ett dt−∫dtdt∫(ett )(dt)2\int {{e^t}{t^2}\,dt = } \,t\int {{e^t}t\,dt - \int {\dfrac{{dt}}{{dt}}{{\int {\left( {{e^t}t\,} \right)\left( {dt} \right)} }^2}} } . . . . . . . . . . . . . (ii)
Now, we will put the value of ∫ett dt\int {{e^t}t\,dt} from (i) into (ii). Thus, we will get following equation:
∫ett2 dt= t[tet−et]−∫(tet−et) dt+c2∫ett2 dt= t2et−tet−∫tetdt+∫etdt+c2 ∫ett2 dt= t2et−tet−(tet−et) +et+c2\begin{array}{l}\int {{e^t}{t^2}\,dt = } \,t\left[ {t{e^t} - {e^t}} \right] - \int {\left( {t{e^t} - {e^t}} \right)} \,dt + {c_2}\\\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \int {t{e^t}dt + \int {{e^t}} dt + {c_2}} \,\\\\\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - t{e^t} - \left( {t{e^t} - {e^t}} \right)\, + {e^t} + {c_2}\end{array}
∫ett2 dt= t2et−2tet +2et+c2\int {{e^t}{t^2}\,dt = } \,{t^2}{e^t} - 2t{e^t}\, + 2{e^t} + {c_2} . . . . . . . . . . . . . (iii)

Now, we will calculate the value of ∫ett3\int {{e^t}{t^3}} by the help of by-parts. In this case, we will take u=t  and  v=ett2u = t\,\,and\,\,v = {e^t}{t^2} . Thus, we will get:
∫ett3 dt= t∫ett2 dt−∫dtdt∫(ett2) (dt)2\int {{e^t}{t^3}\,dt = } \,t\int {{e^t}{t^2}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^2}} \right)} } \,{\left( {dt} \right)^2} . . . . . . . . . . . . . (iv)

Now, we will put the value of ett2{e^t}{t^2} from (iii) to (iv). Thus, we will get:

⇒∫ett3 dt= t[t2et−2tet+2et]−∫(t2et−2tet+2et) dt+c3 ⇒∫ett3 dt= t3et−2t2et+2tet−∫t2etdt+2∫tetdt−2∫etdt +c3 ⇒∫ett3 dt= t3et−2t2et+2tet−[t2et−2tet+2et]+2[tet−et]−2et+c3 ⇒∫ett3 dt= t3et−2t2et+2tet−t2et+2tet−2et+2tet−2et−2et+c3\begin{array}{l} \Rightarrow \int {{e^t}{t^3}\,dt = } \,t\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - \int {\left( {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right)} \,dt + {c_3}\\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \int {{t^2}{e^t}dt + 2\int {t{e^t}dt} - 2\int {{e^t}dt} \,} + {c_3}\\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - \left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] + 2\left[ {t{e^t} - {e^t}} \right] - 2{e^t} + {c_3}\\\ \Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 2{t^2}{e^t} + 2t{e^t} - {t^2}{e^t} + 2t{e^t} - 2{e^t} + 2t{e^t} - 2{e^t} - 2{e^t} + {c_3}\end{array}
⇒∫ett3 dt= t3et−3t2et−6tet−6et+c3\Rightarrow \int {{e^t}{t^3}\,dt = } \,{t^3}{e^t} - 3{t^2}{e^t} - 6t{e^t} - 6{e^t} + {c_3} . . . . . . . . . . . . . (v)

Now, we will calculate the value of ett4dt{e^t}{t^4}dt by the help of by-parts. In this case, we will take u=t  and  v=ett3u = t\,\,and\,\,v = {e^t}{t^3} . Thus, we will get:
∫ett4 dt= t∫ett3 dt−∫dtdt∫(ett3) (dt)2\int {{e^t}{t^4}\,dt = } \,t\int {{e^t}{t^3}\,dt} - \int {\dfrac{{dt}}{{dt}}\int {\left( {{e^t}{t^3}} \right)} } \,{\left( {dt} \right)^2} . . . . . . . . . . . . (vi)
Now, we will put the value of ett3 dt{e^t}{t^3}\,dt from (v) to (vi). Thus, we will get:
\begin{array}{l} \Rightarrow \int {{e^t}{t^4}\,dt = } \,t\left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] - \int {\left( {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right)} \,dt + {c_4}\\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \int {{t^3}{e^t}dt + \int {3{t^2}{e^t}dt} - 6\int {t{e^t}dt} \,} + 6\int {{e^t}dt} \, + {c_4}\\\ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 3{t^3}{e^t} + 6{t^2}{e^t} - 6t{e^t} - \left[ {{t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t}} \right] + 3\left[ {{t^2}{e^t} - 2t{e^t} + 2{e^t}} \right] - 6\left[ {t{e^t} - {e^t}} \right] + 6{e^t} + {c_4}\end{array}$$$$ \Rightarrow \int {{e^t}{t^4}\,dt = } \,{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t} + {c_4} . . . . . . . . . . . . (vii)
Now, we will put the value of ∫ett4\int {{e^t}{t^4}} from (vii) to I. Thus, we will get the following equation:
⇒I=125 [t4et−4t3et+12t2et−24tet+24et]+c ⇒I=et32 [t4−4t3+12t2−24t+24]+c\begin{array}{l} \Rightarrow I = \dfrac{1}{{{2^5}}}\,\left[ {{t^4}{e^t} - 4{t^3}{e^t} + 12{t^2}{e^t} - 24t{e^t} + 24{e^t}} \right] + c\\\ \Rightarrow I = \dfrac{{{e^t}}}{{32}}\,\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}
Now, we will put the value of y back to 2x. Thus, we will get the following result:
⇒I=e2x32 [(2x)4−4(2x)3+12(2x)2−24(2x)+24]+c ⇒I=e2x32 [16x4−32x3+48x2−48x+24]+c ⇒I=8e2x32 [2x4−4x3+6x2−6x+3]+c ⇒I=e2x4 [2x4−4x3+6x2−6x+3]+c\begin{array}{l} \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {{{\left( {2x} \right)}^4} - 4{{\left( {2x} \right)}^3} + 12{{\left( {2x} \right)}^2} - 24\left( {2x} \right) + 24} \right] + c\\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{{32}}\,\left[ {16{x^4} - 32{x^3} + 48{x^2} - 48x + 24} \right] + c\\\ \Rightarrow I = \dfrac{{8{e^{2x}}}}{{32}}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\\\ \Rightarrow I = \dfrac{{{e^{2x}}}}{4}\,\left[ {2{x^4} - 4{x^3} + 6{x^2} - 6x + 3} \right] + c\end{array}
Hence, option (a) is correct.

Note: The shortcut method for finding the integrals of ettn{e^t}{t^n}, where n is an integer is given below:
∫ettn = et[tn−ddt(tn)+d2dt2(tn)−d3dt3(tn)+............(−1)ndndtn(tn)]\int {{e^t}{t^n}\, = } \,{e^t}\left[ {{t^n} - \dfrac{d}{{dt}}\left( {{t^n}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^n}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^n}} \right) + ............{{\left( { - 1} \right)}^n}\dfrac{{{d^n}}}{{d{t^n}}}\left( {{t^n}} \right)} \right]
In our case n=4{\text{n}} = {\text{4}} so, we will get:
∫ett4 = et[t4−ddt(t4)+d2dt2(t4)−d3dt3(t4)+d4dt4(t4)]+c∫ett4 = et[t4−4t3+12t2−24t+24]+c\begin{array}{l}\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - \dfrac{d}{{dt}}\left( {{t^4}} \right) + \dfrac{{{d^2}}}{{d{t^2}}}\left( {{t^4}} \right) - \dfrac{{{d^3}}}{{d{t^3}}}\left( {{t^4}} \right) + \dfrac{{{d^4}}}{{d{t^4}}}\left( {{t^4}} \right)} \right] + c\\\\\int {{e^t}{t^4}\, = } \,{e^t}\left[ {{t^4} - 4{t^3} + 12{t^2} - 24t + 24} \right] + c\end{array}