Why is the derivative of ln(t + √(1+t^2)) equal to 1/(√(1+t^2))?

By chain rule, d/dtln(t+√(1+t^2)) = (1 + 2t/(2√(1+t^2)))/(t+√(1+t^2)) = ((√(1+t^2)+t)/(√(1+t^2)))/(t+√(1+t^2)) = 1/(√(1+t^2)).

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Definite Integral of an Inverse Hyperbolic-Type Logarithmic Function via Substitution and Parts

The value of the integral integral from -log_e 2 to log_e 2 of e^x (log_e (e^x + √(1 + e^2x))) dx is equal to

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Alog_e ((2(2+√(5)))/(√(1+√(5)))) - (√(5))/2Blog_e ((√(2)(3-√(5))^2)/(√(1+√(5)))) + (√(5))/2Clog_e (((2+√(5))^2)/(√(1+√(5)))) + (√(5))/2Dlog_e ((√(2)(2+√(5))^2)/(√(1+√(5)))) - (√(5))/2

Step-by-step solutionView

Correct answer

Using the substitution $t = e^{x}$ followed by integration by parts yields $\log_{e} (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$ .

Option analysis

Why each option works or fails

A · log_e ((2(2+√(5)))/(√(1+√(5)))) - (√(5))/2

You miscalculated the coefficient when factoring out power terms inside the logarithm, replacing $\sqrt{2}$ with $2$ and missing an exponent on $(2 + \sqrt{5})$ . Express the combined logarithmic terms carefully: $2 \log (2 + \sqrt{5}) - \frac{1}{2} \log (1 / 2 + \sqrt{5} / 2)$ rewrites as $\log ((2 + \sqrt{5})^{2} / \sqrt{(1 + \sqrt{5}) / 2}) = \log (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}})$ .

B · log_e ((√(2)(3-√(5))^2)/(√(1+√(5)))) + (√(5))/2

You flipped the sign of the algebraic term during integration by parts from $- \int v d u$ and misidentified the conjugate factor of $(2 + \sqrt{5})$ . Integration by parts formula is $\int u d v = u v - \int v d u$ . The evaluation $[\sqrt{1 + t^{2}}]_{1 / 2}^{2}$ gives $\sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$ , which enters with a minus sign.

C · log_e (((2+√(5))^2)/(√(1+√(5)))) + (√(5))/2

You forgot the factor of $2^{- 1 / 2} = 1 / \sqrt{2}$ coming from the lower limit inside the square root and flipped the sign of the evaluated radical term. At the lower limit $t = 1 / 2$ , $\sqrt{1 + t^{2}} = \sqrt{5} / 2$ , meaning $t + \sqrt{1 + t^{2}} = \frac{1 + \sqrt{5}}{2}$ ; the factor of $1 / 2$ in the denominator pulls out as $\sqrt{2}$ in the numerator.

D · log_e ((√(2)(2+√(5))^2)/(√(1+√(5)))) - (√(5))/2

This is the correct value of the definite integral. Perform substitution $t = e^{x}$ , apply integration by parts with $u = \log (t + \sqrt{1 + t^{2}})$ and $d v = d t$ , and evaluate carefully across the limits $t \in [1 / 2, 2]$ .

Reviewed route

Solution

StepWorking

01Given

Integral $I = \int_{- \ln 2}^{\ln 2} e^{x} \ln (e^{x} + \sqrt{1 + e^{2 x}}) d x$ .

02Approach

Substitute $t = e^{x}$ , then $d t = e^{x} d x$ . The limits transform from $x = - \ln 2$ to $t = 1 / 2$ , and $x = \ln 2$ to $t = 2$ . Then evaluate $\int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$ using Integration by Parts.

03Execute

Let $t = e^{x} ⟹ d t = e^{x} d x$ . Lower limit: $t = e^{- \ln 2} = 1 / 2$ . Upper limit: $t = e^{\ln 2} = 2$ . $I = \int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$ . Using Integration by Parts $\int u \cdot 1 d t = u \cdot t - \int t \cdot u^{'} d t$ , where $u = \ln (t + \sqrt{1 + t^{2}})$ and $u^{'} = \frac{1}{\sqrt{1 + t^{2}}}$ : $I = {[t \ln (t + \sqrt{1 + t^{2}})]}_{1 / 2}^{2} - \int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t$ . Evaluating the boundary term: $= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1}{2} + \sqrt{1 + 1 / 4}) = 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2})$ . Evaluating the integral term: $\int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t = {[\sqrt{1 + t^{2}}]}_{1 / 2}^{2} = \sqrt{5} - \sqrt{5 / 4} = \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$ . Combining terms: $I = \ln ((2 + \sqrt{5})^{2}) - \ln (\sqrt{\frac{\sqrt{5} + 1}{2}}) - \frac{\sqrt{5}}{2} = \ln (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$ .

✓Verify

Check that $\frac{d}{d t} \ln (t + \sqrt{1 + t^{2}}) = \frac{1 + t / \sqrt{1 + t^{2}}}{t + \sqrt{1 + t^{2}}} = \frac{1}{\sqrt{1 + t^{2}}}$ , which is correct. The integrand is positive everywhere on $[- \ln 2, \ln 2]$ , and the resulting value is positive (~1.95).

Hints that build this answer step by step

What is the most direct substitution to simplify the integral $I = \int_{- \log 2}^{\log 2} e^{x} \log (e^{x} + \sqrt{1 + e^{2 x}}) d x$ ?
Substitute $t = e^{x}$ , giving $d t = e^{x} d x$ and limits from $1 / 2$ to $2$ .
Applying integration by parts $\int u d v = u v - \int v d u$ to $\int \log (t + \sqrt{1 + t^{2}}) d t$ , what is the antiderivative?
$t \log (t + \sqrt{1 + t^{2}}) - \sqrt{1 + t^{2}}$
Evaluating ${[t \log (t + \sqrt{1 + t^{2}}) - \sqrt{1 + t^{2}}]}_{1 / 2}^{2}$ , what is the exact combined result?
$\log (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the derivative of

\ln (t + \sqrt{1 + t^{2}})

equal to

\frac{1}{\sqrt{1 + t^{2}}}

By chain rule, $\frac{d}{d t} \ln (t + \sqrt{1 + t^{2}}) = \frac{1 + \frac{2 t}{2 \sqrt{1 + t^{2}}}}{t + \sqrt{1 + t^{2}}} = \frac{\frac{\sqrt{1 + t^{2}} + t}{\sqrt{1 + t^{2}}}}{t + \sqrt{1 + t^{2}}} = \frac{1}{\sqrt{1 + t^{2}}}$ .

Answer

Using the substitution $t = e^{x}$ followed by integration by parts yields $\log_{e} (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$ .

Why each option works or fails

A: log_e ((2(2+√(5)))/(√(1+√(5)))) - (√(5))/2 - You miscalculated the coefficient when factoring out power terms inside the logarithm, replacing $\sqrt{2}$ with $2$ and missing an exponent on $(2 + \sqrt{5})$ . Express the combined logarithmic terms carefully: $2 \log (2 + \sqrt{5}) - \frac{1}{2} \log (1 / 2 + \sqrt{5} / 2)$ rewrites as $\log ((2 + \sqrt{5})^{2} / \sqrt{(1 + \sqrt{5}) / 2}) = \log (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}})$ .
B: log_e ((√(2)(3-√(5))^2)/(√(1+√(5)))) + (√(5))/2 - You flipped the sign of the algebraic term during integration by parts from $- \int v d u$ and misidentified the conjugate factor of $(2 + \sqrt{5})$ . Integration by parts formula is $\int u d v = u v - \int v d u$ . The evaluation $[\sqrt{1 + t^{2}}]_{1 / 2}^{2}$ gives $\sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$ , which enters with a minus sign.
C: log_e (((2+√(5))^2)/(√(1+√(5)))) + (√(5))/2 - You forgot the factor of $2^{- 1 / 2} = 1 / \sqrt{2}$ coming from the lower limit inside the square root and flipped the sign of the evaluated radical term. At the lower limit $t = 1 / 2$ , $\sqrt{1 + t^{2}} = \sqrt{5} / 2$ , meaning $t + \sqrt{1 + t^{2}} = \frac{1 + \sqrt{5}}{2}$ ; the factor of $1 / 2$ in the denominator pulls out as $\sqrt{2}$ in the numerator.
D · correct: log_e ((√(2)(2+√(5))^2)/(√(1+√(5)))) - (√(5))/2 - This is the correct value of the definite integral. Perform substitution $t = e^{x}$ , apply integration by parts with $u = \log (t + \sqrt{1 + t^{2}})$ and $d v = d t$ , and evaluate carefully across the limits $t \in [1 / 2, 2]$ .

Step-by-step solution

Given: Integral $I = \int_{- \ln 2}^{\ln 2} e^{x} \ln (e^{x} + \sqrt{1 + e^{2 x}}) d x$ .
Approach: Substitute $t = e^{x}$ , then $d t = e^{x} d x$ . The limits transform from $x = - \ln 2$ to $t = 1 / 2$ , and $x = \ln 2$ to $t = 2$ . Then evaluate $\int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$ using Integration by Parts.
Execute: Let $t = e^{x} ⟹ d t = e^{x} d x$ . Lower limit: $t = e^{- \ln 2} = 1 / 2$ . Upper limit: $t = e^{\ln 2} = 2$ . $I = \int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$ . Using Integration by Parts $\int u \cdot 1 d t = u \cdot t - \int t \cdot u^{'} d t$ , where $u = \ln (t + \sqrt{1 + t^{2}})$ and $u^{'} = \frac{1}{\sqrt{1 + t^{2}}}$ : $I = {[t \ln (t + \sqrt{1 + t^{2}})]}_{1 / 2}^{2} - \int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t$ . Evaluating the boundary term: $= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1}{2} + \sqrt{1 + 1 / 4}) = 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2})$ . Evaluating the integral term: $\int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t = {[\sqrt{1 + t^{2}}]}_{1 / 2}^{2} = \sqrt{5} - \sqrt{5 / 4} = \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$ . Combining terms: $I = \ln ((2 + \sqrt{5})^{2}) - \ln (\sqrt{\frac{\sqrt{5} + 1}{2}}) - \frac{\sqrt{5}}{2} = \ln (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$ .
Verify: Check that $\frac{d}{d t} \ln (t + \sqrt{1 + t^{2}}) = \frac{1 + t / \sqrt{1 + t^{2}}}{t + \sqrt{1 + t^{2}}} = \frac{1}{\sqrt{1 + t^{2}}}$ , which is correct. The integrand is positive everywhere on $[- \ln 2, \ln 2]$ , and the resulting value is positive (~1.95).

Definite Integral of an Inverse Hyperbolic-Type Logarithmic Function via Substitution and Parts

What feels right?

Using the substitution $t = e^{x}$ followed by integration by parts yields $\log_{e} (\frac{\sqrt{2} (2 + \sqrt{5})^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

✓Verify

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