CBSE Class 12 Maths Exemplar Chapter 7: Integrals NCERT Solutions

Let the integral be denoted by $I$. We have:

I = \int \frac{2x-1}{2x+3} dx

To simplify the integrand, we can rewrite the numerator in terms of the denominator:

I = \int \frac{(2x+3) - 3 - 1}{2x+3} dx = \int \frac{(2x+3) - 4}{2x+3} dx

Separating the terms, we get:

I = \int \left( \frac{2x+3}{2x+3} - \frac{4}{2x+3} \right) dx = \int \left( 1 - \frac{4}{2x+3} \right) dx

Now, we integrate term by term:

I = \int 1 dx - \int \frac{4}{2x+3} dx

The first integral is $x$. For the second integral, we can use a simple substitution or recognize the form $\int \frac{a}{bx+c} dx$. Let $u = 2x+3$, so $du = 2dx$, which means $dx = \frac{1}{2}du$. The integral becomes:

\int \frac{4}{u} \left( \frac{1}{2} du \right) = 2 \int \frac{1}{u} du = 2 \log|u| = 2 \log|2x+3|

Combining these results:

I = x - 2 \log|2x+3| + C'

We can use the property of logarithms $n \log m = \log m^n$ to rewrite $2 \log|2x+3|$ as $\log|(2x+3)^2|$.

I = x - \log|(2x+3)^2| + C

where $C = C'$ is the constant of integration. Thus, the given integral is verified.

Let the integral be $I$.

I = \int \frac{2x+3}{x^2+3x} dx

We can solve this integral using the substitution method. Let $t = x^2+3x$. Then, the differential $dt$ is given by:

dt = (2x+3) dx

Substituting $t$ and $dt$ into the integral, we get:

I = \int \frac{1}{t} dt

The integral of $\frac{1}{t}$ with respect to $t$ is $\log|t|$.

I = \log|t| + C

Substituting back $t = x^2+3x$, we obtain:

I = \log|x^2+3x| + C

This verifies the given integral statement.

Let the integral be $I$.

I = \int \frac{x^2+2}{x+1} dx

We can perform polynomial long division or algebraic manipulation to simplify the integrand. Let's use algebraic manipulation:

I = \int \frac{x^2 - 1 + 1 + 2}{x+1} dx = \int \frac{(x^2-1) + 3}{x+1} dx

Using the difference of squares formula ($x^2-1 = (x-1)(x+1)$):

I = \int \frac{(x-1)(x+1) + 3}{x+1} dx = \int \left( \frac{(x-1)(x+1)}{x+1} + \frac{3}{x+1} \right) dx

Simplifying the expression:

I = \int \left( x-1 + \frac{3}{x+1} \right) dx

Now, we integrate term by term:

I = \int (x-1) dx + \int \frac{3}{x+1} dx

Integrating $x-1$ gives $\frac{x^2}{2} - x$. Integrating $\frac{3}{x+1}$ gives $3 \log|x+1|$.

I = \frac{x^2}{2} - x + 3 \log|x+1| + C

Therefore, the value of the integral is $\frac{x^2}{2} - x + 3 \log|x+1| + C$.

Let the integral be $I$.

I = \int \frac{e^{6\log x} - e^{5\log x}}{e^{4\log x} - e^{3\log x}} dx

First, we use the logarithmic property $a \log b = \log b^a$ to rewrite the exponents:

e^{6\log x} = e^{\log x^6} = x^6

e^{5\log x} = e^{\log x^5} = x^5

e^{4\log x} = e^{\log x^4} = x^4

e^{3\log x} = e^{\log x^3} = x^3

Substitute these back into the integral:

I = \int \frac{x^6 - x^5}{x^4 - x^3} dx

Now, factor out common terms from the numerator and the denominator:

I = \int \frac{x^5(x - 1)}{x^3(x - 1)} dx

Cancel out the common factor $(x-1)$, assuming $x \neq 1$:

I = \int \frac{x^5}{x^3} dx = \int x^{5-3} dx = \int x^2 dx

Finally, integrate $x^2$ using the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$):

I = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C

Thus, the value of the integral is $\frac{x^3}{3} + C$.

Let the integral be $I$.

I = \int \frac{(1 + \cos x)}{(x + \sin x)} dx

We can solve this integral using the substitution method. Let $t = x + \sin x$. Then, the differential $dt$ is found by differentiating $t$ with respect to $x$:

\frac{dt}{dx} = \frac{d}{dx}(x + \sin x) = 1 + \cos x

This implies that:

dt = (1 + \cos x) dx

Now, substitute $t$ and $dt$ into the integral:

I = \int \frac{1}{t} dt

The integral of $\frac{1}{t}$ with respect to $t$ is $\log|t|$.

I = \log|t| + C

Substitute back $t = x + \sin x$ to get the final answer in terms of $x$:

I = \log|x + \sin x| + C

Therefore, the value of the integral is $\log|x + \sin x| + C$.