COMPASSLearning

Talk to us Register Login

CBSE Class 12 · 2014 · Definite IntegralsLong Answer

Evaluate:

\int_{2}^{4} \frac{x}{x ^{2} + 1} d x

Answer: $\frac{1}{2} lo g \frac{17}{5}$

Solution

Put

t = x^{2} + 1 \Rightarrow d t = 2 x d x

, so

x d x = \frac{1}{2} d t

. Limits:

x = 2 \to t = 5

,

x = 4 \to t = 17

.

I = \frac{1}{2} \int_{5}^{17} \frac{d t}{t} = \frac{1}{2} [lo g ∣ t ∣]_{5}^{17} = \frac{1}{2} (lo g 17 - lo g 5) = \frac{1}{2} lo g \frac{17}{5}

CBSE Class 12 Board · 2014 · Subjective

Practise more: all CBSE board PYQs on Definite Integrals →

More board questions on Definite Integrals

The value of $\int_{π /4}^{π /2} cot θ cosec^{2} θ d θ$ is
CBSE 2024View solution →
Evaluate: $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$
CBSE 2024View solution →
Evaluate $\int_{l o g 2}^{l o g 3} \frac{1}{( e ^{x} + e ^{- x} ) ( e ^{x} - e ^{- x} )} d x$
CBSE 2023View solution →
Evaluate $\int_{- 1}^{1} x^{4} - x d x$
CBSE 2023View solution →
Evaluate $\int_{0}^{π /2} \frac{1}{1 + ( t a n x ) ^{2/3}} d x$
CBSE 2022View solution →
$\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{2} x d x$ :
CBSE 2020View solution →