MHTCET 26 Apr 2025 Shift 1 Paper

Question : 117 of 150

Marks: +1, -0

\int_{0}^{1} \tan ^{-1} x d x=

Solution:

Concept:

Use integration by parts to evaluate

\int \tan^{-1} x \, dx

Explanation:

Let

I = \int_{0}^{1} \tan^{-1} x \, dx

Take

\tan^{-1} x

as the first function and

1

as the second function.

Using integration by parts:

I = \left[x \tan^{-1} x\right]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x^2} \, dx

I = \left[x \tan^{-1} x - \frac{1}{2}\log(1+x^2)\right]_{0}^{1}

Substitute the limits:

I = \left(1 \cdot \frac{\pi}{4} - \frac{1}{2}\log 2\right) - \left(0 - 0\right)

I = \frac{\pi}{4} - \frac{1}{2}\log 2

Since

\frac{1}{2}\log 2 = \log \sqrt{2}

, we get:

I = \frac{\pi}{4} - \log \sqrt{2}

Answer:

Option B:

\frac{\pi}{4}-\log \sqrt{2}

Go to Question:

Prev Question

Next Question