Question

Given, \[\int {(1 + x - {x^{ - 1}})} \]\[{e^{x + {x^{ - 1}}}}\] \[dx\] equals
A.\[(1 + x){e^{x + {x^{ - 1}}}} + C\]
B.\[(x - 1){e^{x + {x^{ - 1}}}} + C\]
C.\[ - x{e^{x + {x^{ - 1}}}} + C\]
D.\[x{e^{x + {x^{ - 1}}}} + C\]

Answer 1

Hint: In this, we will first open the brackets and multiply \[{e^{x + {x^{ - 1}}}}\] with all the terms in the bracket. After that, we see that \[\dfrac{d}{{dx}}(x + {x^{ - 1}}) = 1 - {x^{ - 1 - 1}} = 1 - {x^{ - 2}}\] and then we try to write \[x - {x^{ - 1}}\] as a product of \[1 - {x^{ - 2}}\] and some other term. Then, after that substitute \[x + {x^1} = t\] and then apply By Parts to the second integral, which will result in cancellation of the first integral with one of the terms in the obtained value of the second integral.
By Parts Formula –
\[\int {uv} \]\[dx\]\[ = u \times \int {(v)} dx - \int {[(\dfrac{d}{{dx}}u) \times \int {(v)} dx} ]dx\]
Where, u is First function and v is second function which are decided by ILATE, where
I stands for Inverse Trigonometry Function
L stands for Logarithmic Function
A stands for Algebraic Function
T stands for Trigonometric Function
E stands for Exponential Function

Complete answer:
\[\int {(1 + x - {x^{ - 1}})} \]\[{e^{x + {x^{ - 1}}}}\]\[dx\]\[ = \int {({e^{x + {x^{ - 1}}}} + (x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}} )dx\]
\[ = \int {({e^{x + {x^{ - 1}}}})dx + \int {((x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}} } )dx\] (\[\int {(a + b)dx = \int {(a)dx} } + \int {(b)dx} \])
 \[ = \int {({e^{x + {x^{ - 1}}}})dx + \int {(x(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}} } )dx\] (Taking \[x\] common from \[x - {x^{ - 1}}\]) -(1)
Now, Taking the second Integral and solving using By Parts
\[ = \int {x(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}} dx\]
Now, taking \[x\]as first function and \[(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}\] as second function, we get
\[\int {x(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}} dx = x \times \int {(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}dx} - \int {[\dfrac{d}{{dx}}x \times \int {(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}dx]dx} } \] ----(2)
For solving \[\int {(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}dx} \], we substitute
 \[x + {x^{ - 1}} = t\] ----(3)
Now, differentiating the above substitution with respect to \[x\], we get
\[(1 - {x^{ - 2}})dx = dt\] ----(4)
Using (3) and (4) in (2), we get
\[\int {x(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}} dx = x \times \int {{e^t}dt} - \int {[\dfrac{d}{{dx}}x \times \int {{e^t}dt]} dx} \]
 \[ = x \times {e^t} - \int {(1 \times {e^t})dx} + c\] (\[\int {{e^x}} dx = {e^x} + c\]
                                                                                                                            (\[\dfrac{d}{{dx}}x = 1\])
\[ = x{e^t} - \int {{e^t}} dx + c\]
Now, replacing back the value of \[t\] in terms of \[x\] from (3) in the above equation
\[\int {x(1 - {x^{ - 2}}){e^{x + {x^{ - 1}}}}} dx = x{e^{x + {x^{ - 1}}}} - \int {{e^{x + {x^{ - 1}}}}} dx + c\] -----(5)
Substituting (5) in (1), we get
\[\int {(1 + x - {x^{ - 1}})} \]\[{e^{x + {x^{ - 1}}}}dx\]\[ = \int {{e^{x + {x^{ - 1}}}}dx + x{e^{x + {x^{ - 1}}}} - \int {{e^{x + {x^{ - 1}}}}} } dx + c\]
\[ = x{e^{x + {x^{ - 1}}}} + c\]
Hence, we get
\[\int {(1 + x - {x^{ - 1}})} {e^{x + {x^{ - 1}}}}dx = x{e^{x + {x^{ - 1}}}} + c\]
Therefore, the correct option is d.

Note:
Firstly, we need to be very careful about the constant while we solve the integration problem. We usually forget to add the constant after solving the integral. Also, we need to take care which function we should take as first function and which function we should choose as second function while applying By Parts in the question. Also, if we know that some of the terms will get cancelled without solving them , we can keep them as they are and then at the end we can cancel them out.