Definite IntegrationhardFree

∫ eᵗ/(t+a)² dt in terms of f(x)=∫eᵗ/t² dt | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

f(x)=\displaystyle\int_{1}^{x}\dfrac{e^{t}}{t^{2}}\,dt

for all

x>0

. Then the value of

\displaystyle\int_{1}^{x}\dfrac{e^{t}}{(t+a)^{2}}\,dt

e^{-a}\big(f(x+a)+f(1+a)\big)

e^{-a}\big(f(x+a)+f(a)\big)

e^{-a}\big(f(x+a)-f(1+a)\big)

correct

DNone of these

Solution

Step 1: Substitute

t+a=z

, so

t=z-a

and

dt=dz

. The limits map as

t=1\Rightarrow z=1+a

and

t=x\Rightarrow z=x+a

\int_{1}^{x}\dfrac{e^{t}}{(t+a)^{2}}\,dt=\int_{1+a}^{x+a}\dfrac{e^{z-a}}{z^{2}}\,dz.

Step 2: Factor out

e^{-a}

, which is constant with respect to

z

=e^{-a}\int_{1+a}^{x+a}\dfrac{e^{z}}{z^{2}}\,dz.

Step 3: Recognise the integrand as the one defining

f

. Since

f(u)=\displaystyle\int_{1}^{u}\dfrac{e^{t}}{t^{2}}\,dt

, the integral over

[1+a,\,x+a]

is a difference of

f

-values:

\int_{1+a}^{x+a}\dfrac{e^{z}}{z^{2}}\,dz=f(x+a)-f(1+a).

Step 4: Combine:

\int_{1}^{x}\dfrac{e^{t}}{(t+a)^{2}}\,dt=e^{-a}\big(f(x+a)-f(1+a)\big).

Correct answer: (3)

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