Verify: $\int \dfrac{2x-1}{2x+3}\,dx = x - \log|(2x+3)^2| + C$

> [!TLDR] > **True.** Differentiating $x-\log|(2x+3)^2|+C$ gives $\dfrac{2x-1}{2x+3}$, so the stated antiderivative is correct. The fastest way to *verify* a claimed integral is to differentiate the proposed answer: if you recover the integrand, the statement holds. No integration is needed. ## Differentiate the right-hand side Let $F(x)=x-\log|(2x+3)^2|+C$. First simplify the logarithm with the power rule $\log|a^2|=2\log|a|$: $$F(x)=x-2\log|2x+3|+C.$$ Now differentiate term by term: - $\dfrac{d}{dx}(x)=1$, - $\dfrac{d}{dx}\big(-2\log|2x+3|\big)=-2\cdot\dfrac{1}{2x+3}\cdot 2=-\dfrac{4}{2x+3}.$ So $$F'(x)=1-\frac{4}{2x+3}=\frac{(2x+3)-4}{2x+3}=\frac{2x-1}{2x+3}.$$ ## Compare with the integrand This matches $\dfrac{2x-1}{2x+3}$ exactly, and the domains agree ($x\neq-\tfrac32$). Hence $F$ is a valid antiderivative and the identity is correct. > [!NOTE] > Writing the constant as $\log|(2x+3)^2|$ instead of $2\log|2x+3|$ is just a stylistic choice — the two are equal, so both forms verify identically. > [!ANSWER] > **True.** $\dfrac{d}{dx}\big[x-\log|(2x+3)^2|\big]=\dfrac{2x-1}{2x+3}$, confirming $\displaystyle\int\frac{2x-1}{2x+3}\,dx=x-\log|(2x+3)^2|+C.$

Verify: ∫ (2x-1)/(2x+3)dx = x - log|(2x+3)²| + C – Class 12 Mathematics

NCERT Exemplar · Q1

\int \dfrac{2x-1}{2x+3}\,dx = x - \log|(2x+3)^2| + C

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Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

A function $y = \phi(x)$ is called a solution of a differential equation if, when you substitute it and its derivatives into the equation, the two sides become equal for every $x$ in the domain. Verification is the act of carrying out that substitution and checking that it holds as an identity.

The useful point: you do not have to solve the equation to verify a candidate. You are only checking a function that is already handed to you — which is exactly how many exam questions are phrased: "Show that $\dots$ is a solution of $\dots$ ."

The steps

From the given $y = \phi(x)$ , compute exactly the derivatives that appear in the equation.
Substitute $y$ and those derivatives into the left-hand side.
Simplify and check that it equals the right-hand side for all $x$ (an identity, not just at one point).

Example 1

Verify that $y = e^{-3x}$ is a solution of $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$ .

Here $y' = -3e^{-3x}$ and $y'' = 9e^{-3x}$ . Substituting:

9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = (9 - 3 - 6)e^{-3x} = 0.

The left side is $0$ for every $x$ , so $y = e^{-3x}$ is a solution.

Example 2 (a solution with constants)

Verify that $y = a\cos x + b\sin x$ satisfies $\dfrac{d^2y}{dx^2} + y = 0$ for any constants $a, b$ .

Since $y'' = -a\cos x - b\sin x = -y$ , we get $y'' + y = 0$ . It holds for all $a, b$ , so this two-constant family is a solution.

Note

Verification links your answer back to the definition of a solution: a function is a solution not because of how you found it, but because it makes the differential equation true. If the substitution does not reduce to an identity, the function is simply not a solution.

Verifying that a given function solves a differential equation is explicitly listed as an exercise type in the NCERT Class 12 Mathematics textbook's Differential Equations chapter, and "verify the solution of differential equation examples" is a common CBSE and JEE Main search. This is often the easiest full-mark question in the chapter once the substitution steps are practiced a few times.

True. Differentiating $x-\log|(2x+3)^2|+C$ gives $\dfrac{2x-1}{2x+3}$ , so the stated antiderivative is correct.

The fastest way to verify a claimed integral is to differentiate the proposed answer: if you recover the integrand, the statement holds. No integration is needed.

Differentiate the right-hand side

Let $F(x)=x-\log|(2x+3)^2|+C$ . First simplify the logarithm with the power rule $\log|a^2|=2\log|a|$ :

F(x)=x-2\log|2x+3|+C.

Now differentiate term by term:

$\dfrac{d}{dx}(x)=1$ ,
$\dfrac{d}{dx}\big(-2\log|2x+3|\big)=-2\cdot\dfrac{1}{2x+3}\cdot 2=-\dfrac{4}{2x+3}.$

F'(x)=1-\frac{4}{2x+3}=\frac{(2x+3)-4}{2x+3}=\frac{2x-1}{2x+3}.

Compare with the integrand

This matches $\dfrac{2x-1}{2x+3}$ exactly, and the domains agree ( $x\neq-\tfrac32$ ). Hence $F$ is a valid antiderivative and the identity is correct.

Note

Writing the constant as $\log|(2x+3)^2|$ instead of $2\log|2x+3|$ is just a stylistic choice — the two are equal, so both forms verify identically.

✓Final answer

\dfrac{d}{dx}\big[x-\log|(2x+3)^2|\big]=\dfrac{2x-1}{2x+3}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL4 marks
$\left(\dfrac{d^2y}{dx^2}\right)^5 + \left(\dfrac{dy}{dx}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) + 1 = 0$
›Reveal solutionSolution

The ODE has order 2 but no defined degree (it isn't a polynomial in derivatives); $y=\sqrt{1+x^2}$ checks out as a genuine solution of the given first-order ODE. The OR part solves a separable ODE with an initial condition.

Order and degree: In $\left(\dfrac{d^2y}{dx^2}\right)^5+\left(\dfrac{dy}{dx}\right)^2+\cos\left(\dfrac{dy}{dx}\right)+1=0$ , the highest-order derivative is $d^2y/dx^2$ , so the order is 2. Degree is defined only when the equation is a polynomial in the derivatives; here $\cos(dy/dx)$ is a transcendental function of $dy/dx$ , not a polynomial term, so the degree is not defined.

Verification: $y=\sqrt{1+x^2}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+x^2}}$ . The RHS of the given equation is

$\frac{xy}{1+x^2}=\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}=\frac{dy}{dx}.$

Since LHS $=$ RHS, $y=\sqrt{1+x^2}$ is indeed a solution.

OR: $x(x^2-1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x(x-1)(x+1)}$ . Partial fractions:

$\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1},$

giving (by substitution) $A=-1,\ B=\tfrac12,\ C=\tfrac12$ . So

$y=-\ln|x|+\frac12\ln|x-1|+\frac12\ln|x+1|+K=-\ln|x|+\frac12\ln|x^2-1|+K.$

Using $y=0$ at $x=2$ : $0=-\ln2+\tfrac12\ln3+K\Rightarrow K=\ln2-\tfrac12\ln3$ . So the particular solution is

$y=-\ln x+\frac12\ln(x^2-1)+\ln2-\frac12\ln3=\ln\frac{2}{x}+\frac12\ln\frac{x^2-1}{3}\quad(x>2\text{ region}).$

✓Final answer
$=2$
AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL4 marks
$y = 3\cos(\log x) + 4\sin(\log x)$
›Reveal solutionSolution

Differentiate twice using the chain rule, multiply by $x$ and $x^2$ appropriately, and the terms cancel to give $0$ .

$y = 3\cos(\log x) + 4\sin(\log x)$ .

First derivative (chain rule, $\frac{d}{dx}\log x = 1/x$ ):

$y' = -3\sin(\log x)\cdot\frac1x + 4\cos(\log x)\cdot\frac1x = \frac1x\left[4\cos(\log x) - 3\sin(\log x)\right].$

So $xy' = 4\cos(\log x) - 3\sin(\log x)$ .

Differentiate $xy'$ again with respect to $x$ (product rule on the LHS, chain rule on the RHS):

$y' + xy'' = -4\sin(\log x)\cdot\frac1x - 3\cos(\log x)\cdot\frac1x = -\frac1x\left[3\cos(\log x)+4\sin(\log x)\right] = -\frac{y}{x}.$

Multiply both sides by $x$ :

$xy' + x^2y'' = -y.$

Rearranging:

$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0.$

This is exactly the required relation, so it is proved.

✓Final answer
$x^2y''+xy'+y=0$

←Previous section · summarySummary →Next questionQ2 — Verify: ∫ (2x+3)/(x²+3x)dx = log|x²+3x| + C

Q.Verify: $\int \dfrac{2x-1}{2x+3}\,dx = x - \log|(2x+3)^2| + C$

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

The steps

Example 1

Example 2 (a solution with constants)

Differentiate the right-hand side

Compare with the integrand

Method: Verifying a claimed antiderivative by differentiation

Steps

Common Mistakes

Q.Verify: ∫2x−12x+3 dx=x−log⁡∣(2x+3)2∣+C\int \dfrac{2x-1}{2x+3}\,dx = x - \log|(2x+3)^2| + C∫2x+32x−1​dx=x−log∣(2x+3)2∣+C

Differentiate the right-hand side

Compare with the integrand

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Q.Verify: $\int \dfrac{2x-1}{2x+3}\,dx = x - \log|(2x+3)^2| + C$

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