Question

If $\int e^{2x}.x^{4}dx=\frac{e^{2x}}{2}f\left(x\right)+C$ then f(x) is equal to

• A. $(x^4-2x^3+3x^2-3x+\frac{3}{2})\frac{1}{2}$
• B. $x^4-x^3+2x^2-3x+2$
• C. $x^4-2x^3+3x^2-3x+\frac{3}{2}$
• D. $x^4-2x^3+2x^3-3x+\frac{3}{2}$

Answer 1

The correct option is C $x^4-2x^3+3x^2-3x+\frac{3}{2}$

$I=\int e^{2x}.x^{4}dx$
Applying integration by parts,
$=x^4\frac{e^{2x}}{2}-\int {4x}^3\frac{e^{2x}}{2}dx$
$I=x^4\frac{e^{2x}}{2}-2\int x^3{e}^{2x}dx$
Again applying integration by parts
$I=x^4\frac{e^{2x}}{2}-2x^3\frac{e^{2x}}{2}+2\int {3x}^2\frac{e^{2x}}{2}dx$
$\Rightarrow I=x^4\frac{e^{2x}}{2}-x^3{e}^{2x}+3\int x^2{e}^{2x}dx$
Again applying integration by parts, we get
$I=x^4\frac{e^{2x}}{2}-x^3{e}^{2x}+3x^2\frac{e^{2x}}{2}-3\int 2x\frac{e^{2x}}{2}dx$
$I=x^4\frac{e^{2x}}{2}-x^3{e}^{2x}+\frac{3}{2}{x}^2{e}^{2x}-3\int x{e}^{2x}dx$
Again applying integration by parts, we get
$I=x^4\frac{e^{2x}}{2}-x^3{e}^{2x}+\frac{3}{2}{x}^2{e}^{2x}-\frac{3}{2}x{e}^{2x}+3\int \frac{e^{2x}}{2}dx$
$\Rightarrow I=x^4\frac{e^{2x}}{2}-x^3{e}^{2x}+\frac{3}{2}{x}^2{e}^{2x}-\frac{3}{2}x{e}^{2x}+\frac{3}{4}{e}^{2x}+C$
$\Rightarrow I=\frac{e^{2x}}{2}(x^4-2x^3+3x^2-3x+\frac{3}{2})+C$
On comparing with given , we get
$f\left(x\right)=x^4-2x^3+3x^2-3x+\frac{3}{2}$