Question

$\int \frac{e^{2x}dx}{\sqrt[4]{4e^x-1}}$ is equal to

• A. $\frac{4}{21}(e^x-1{)}^{3/4}(3e^x+4)+k$
• B. $\frac{(e^x-1{)}^{1/4}(3e^x+4)}{21}+k$
• C. $\frac{4}{21}(e^x+1)(3e^x+4)+k$
• D. $(e^x-1{)}^{3/4}+k$

Answer 1

The correct option is A $\frac{4}{21}(e^x-1{)}^{3/4}(3e^x+4)+k$

$(e^x-1{)}^{1/4}=t$

$e^{2x}=(t^4+1{)}^2$

$2.e^{2x}.dx=2(t^4+1)4t^{3}dt$

$\int \frac{(t^4+1{)}^2(4t^3)dt}{t}$

$4\int (t^4+1{)}^2(t^2)dt$

$4\int (t^8+2t^4+1)t^{2}dt$

$4\int (t^{10}+2t^6+t^2)dt$

$4[\frac{t^{11}}{11}+\frac{2t^7}{7}+\frac{t^3}{3}]$

$4t^3[\frac{t^8}{11}+\frac{2t^4}{7}+\frac{1}{3}]+c$

$4(e^x-1{)}^{3/4}[\frac{(e^x-1{)}^2}{11}+\frac{2(e^x-1)}{4}+\frac{1}{3}]$

$=4/21(e^x-1{)}^{3/4}[3e^x+4]+k$