Use the substitution $u = 2^x$ to find the exact value of
\[ \int_0^1 \frac{2^x}{(2^x + 1)^2} \, dx - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 7
Question 4
Use the substitution $u = 2^x$ to find the exact value of
\[ \int_0^1 \frac{2^x}{(2^x + 1)^2} \, dx. \]
Worked Solution & Example Answer:Use the substitution $u = 2^x$ to find the exact value of
\[ \int_0^1 \frac{2^x}{(2^x + 1)^2} \, dx - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 7
Step 1
Substitution: $u = 2^x$
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Answer
We start by substituting ( u = 2^x ). Then we find the differential:
[ \frac{du}{dx} = 2^x \ln{2} \Rightarrow dx = \frac{du}{2^x \ln{2}} = \frac{du}{u \ln{2}}. ]
We also need to change the limits of integration. For ( x = 0 ):
[ u = 2^0 = 1. ]
For ( x = 1 ):
[ u = 2^1 = 2. ]
Step 2
Rewrite the Integral
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Answer
Now, substituting into the integral, we have:
[ \int_1^2 \frac{u}{(u + 1)^2} \cdot \frac{du}{u \ln{2}} = \frac{1}{\ln{2}} \int_1^2 \frac{1}{(u + 1)^2} , du. ]
Step 3
Evaluate the Integral
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Answer
To evaluate ( \int \frac{1}{(u + 1)^2} , du ), we recognize it as:
[ -\frac{1}{u + 1}. ]
Therefore,
[ \int_1^2 \frac{1}{(u + 1)^2} , du = -\left[ \frac{1}{u + 1} \right]_1^2 = -\left( \frac{1}{3} - \frac{1}{2} \right) = -\left( -\frac{1}{6} \right) = \frac{1}{6}. ]
Step 4
Combine Results
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Answer
Putting this together, we get:
[ \frac{1}{\ln{2}} \cdot \frac{1}{6} = \frac{1}{6 \ln{2}}. ]
Thus, the exact value of the integral is:
[ \frac{1}{6 \ln{2}}. ]
